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questions</a></label></li></ul></div> <button id="show-periodic-table" class="btn btn-default btn-block" style="margin-bottom: 10px"><i class="fa fa-table"></i> Periodic table</button> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Stoichiometry core (SL & HL) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2123" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../2075/paper-1-exam-questions.html">Paper 1 Exam Questions</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Stoichiometry core (SL & HL) paper 1 questions</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topic 1 (1.1, 1.2 and 1.3)</strong></p> <p><strong>Paper 1 style questions </strong>are multiple choice. You are <strong>not permitted to use a calculator or the data book</strong> for these questions, but you should use a periodic table.</p> <p>A <strong>periodic table pop-up</strong> is available on the left hand menu.</p> <div class="greenBg"> <div class="tib-quiz" data-stats="7-585-2123"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>0.20 mol of hydrochloric acid is mixed with 0.20 mol of calcium carbonate:</p><p>2HCl + CaCO<sub>3</sub> → CaCl<sub>2</sub> + H<sub>2</sub>O + CO<sub>2</sub></p><p>Which is correct?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> Limiting reagent is CaCO<sub>3</sub> and maximum yield of CO<sub>2</sub> is 0.10 mol</label></p><p><label class="radio"><input type="radio"> Limiting reagent is HCl and maximum yield of CO<sub>2</sub> is 0.20 mol</label></p><p><label class="radio"><input class="c" type="radio"> Limiting reagent is HCl and maximum yield of CO<sub>2</sub> is 0.10 mol</label></p><p><label class="radio"><input type="radio"> Limiting reagent is CaCO<sub>3 </sub>and maximum yield of CO<sub>2</sub> is 0.20 mol</label></p></div><div class="q-explanation"><p>Quantities of both reagents are given, which suggests that one reagent will be in excess and the other will be limiting. This is best considered in terms of the mole ratio:</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;">2HCl</td><td style="text-align: center;">+</td><td style="text-align: center;">CaCO<sub>3</sub></td><td style="text-align: center;">→</td><td style="text-align: center;">CaCl<sub>2</sub></td><td style="text-align: center;">+</td><td style="text-align: center;">H<sub>2</sub>O</td><td style="text-align: center;">+</td><td style="text-align: center;">CO<sub>2</sub></td></tr><tr><td style="text-align: center;">2</td><td style="text-align: center;">:</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">1</td></tr></tbody></table><p>2 moles of HCl will react with 1 mole of CaCO<sub>3</sub>.</p><p>Therefore 0.20 moles of HCl will react with 0.10 moles of CaCO<sub>3</sub>. 0.20 moles of CaCO<sub>3</sub> has been used, so the calcium carbonate is in excess, and the hydrochloric acid is therefore the limiting reagent.</p><p>2 moles of HCl will yield 1 mole of CO<sub>2</sub>.</p><p>Therefore 0.20 moles of HCl will yield 0.10 moles of CO<sub>2</sub>.</p><p>Thus <strong>Limiting reagent is HCl and maximum yield of CO<sub>2</sub> is 0.10 mol </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the sum of all coefficients when the following equation is balanced using the smallest possible whole numbers?</p><p>___GeO<sub>2</sub> <sub> </sub>+ ___HCl → ___GeCl<sub>4</sub> + ___H<sub>2</sub>O</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> 8</label></p><p><label class="radio"> <input type="radio"> 6</label></p><p><label class="radio"> <input type="radio"> <span>4</span></label> </p><p><label class="radio"> <input type="radio"> 7</label></p></div><div class="q-explanation"><p><span style="font-size: 15px;"></span>GeO<sub>2</sub> + 4HCl → GeCl<sub>4</sub> + 2H<sub>2</sub>O<sub> </sub>is the balanced equation, so the total sum is 8 - don't forget that GeO<sub>2</sub> and GeCl<sub>4</sub> both have a '1' in front of them, although by convention we don't actually write that into the equation.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>Which contains the greatest number of moles of oxygen atoms?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> 0.05 mol Ca(NO<sub>3</sub>)<sub>2</sub></label></p><p><label class="radio"><input type="radio"> 0.10 mol O<sub>2</sub></label></p><p><label class="radio"><input type="radio"> 0.20 mol CO</label></p><p><label class="radio"><input class="c" type="radio"> 0.10 mol C<sub>6</sub>H<sub>4</sub>(NO<sub>2</sub>)<sub>2</sub></label></p></div><div class="q-explanation"><p>This question requires us first to work out the number of oxygen atoms in each formula, and to then multiply that by the number of moles, which will give us the number of moles of oxygen atoms:</p><p>0.05 mol Ca(NO<sub>3</sub>)<sub>2 </sub>: calcium nitrate has 6 oxygen atoms: 6 × 0.05 = 0.30</p><p>0.10 mol C<sub>6</sub>H<sub>4</sub>(NO<sub>2</sub>)<sub>2 </sub>: dinitrobenzene has 4 oxygen atoms: 4 × 0.10 = 0.40</p><p>0.20 mol CO : carbon monoxide has 1 oxygen atom: 1 × 0.20 = 0.20</p><p>0.10 mol O<sub>2</sub> : oxygen has 2 oxygen atom: 2 × 0.10 = 0.20</p><p>Thus <strong>0.10 mol C<sub>6</sub>H<sub>4</sub>(NO<sub>2</sub>)<sub>2</sub> </strong>is the correct answer as this contains the largest number of moles of oxygen atoms (0.40 mol).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of these is an example of a heterogeneous mixture?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> Air </label></p><p><label class="radio"> <input type="radio"> A<span>queous sodium chloride</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Solid calcium carbonate powder suspended in water</span></label> </p><p><label class="radio"> <input type="radio"> Nitrogen dioxide gas<span></span></label> </p></div><div class="q-explanation"><p>A heterogeneous mixture is a mixture of more than one phase (or state) of matter. Hence <strong>Solid calcium carbonate powder suspended in water </strong>is the correct answer as this contains both a solid (calcium carbonate powder) and a liquid (water). The other answers are homogeneous (one phase of matter) mixtures (a solution like aqueous sodium chloride is uniform throughout and considered to be homogeneous), or in the case of nitrogen dioxide gas - a pure compound.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What is the empirical formula of a hydrocarbon with 90% carbon and 10% hydrogen by mass?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>C<sub>7</sub>H<sub>10</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>4</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>9</sub>H</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C<sub>3</sub>H<sub>4</sub></span></label> </p></div><div class="q-explanation"><p>We need to find the simplest ratio of the number of (moles of) atoms. We cannot use a calculator so we will need to approximate the relative atomic masses that we take from the periodic table (e.g. carbon 12.01 = 12).</p><p>Start by assuming 100g: Thus 90g of C atoms, 10g of H atoms:</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Carbon</td><td style="text-align: center;">Hydrogen</td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;"><p>moles = mass/molar mass</p></td><td style="text-align: center;">90/12 = 7.5</td><td style="text-align: center;">10/1 = 10</td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;">multiply up to whole numbers (×2)</td><td style="text-align: center;">15</td><td style="text-align: center;">20</td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;">find the largest common multiple to get the simplest ratio (5 in this case)</td><td style="text-align: center;">15 ÷ 5 = <strong>3</strong></td><td style="text-align: center;">20 ÷ 5 =<strong> 4</strong></td><td style="text-align: center;"> </td></tr></tbody></table><p>Therefore the empirical formula and the correct answer is <strong>C<sub>3</sub>H<sub>4</sub></strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which relationship would give the following graph when plotted for a fixed mass of an ideal gas with all other variables constant?</p><p><img alt="" src="../../images/stoichiometry/pvsvgraph.png" style="width: 240px; height: 168px;"></p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio"> P against V</label></p><p><label class="radio"><input type="radio"> P against <span class="math-tex">\({1 \over V}\)</span></label></p><p><label class="radio"><input type="radio"> P against T</label></p><p><label class="radio"><input type="radio"> V against T</label></p></div><div class="q-explanation"><p>The ideal gas equation is PV=nRT. The letters represent Pressure, Volume, number of moles, gas constant (R) and Temperature in that order.</p><p>P against T and V against T will both give a directly proportional linear relationship as the two variables are on opposite sides of the equals sign in the equation; so any change in one variable must be proportionally mirrored in the other variable if all the other values are constant.</p><p>P against <span class="math-tex">\({1\over V}\)</span> will also give a straight line. If all other values in PV=nRT are constant then PV=constant. If we give that constant a value of 1 (PV=1) then P =<span class="math-tex">\({1 \over V}\)</span> so volume is inversely proportional to pressure and P against <span class="math-tex">\({1\over V}\)</span> will give a straight line when plotted.</p><p>P against V will give the curve (a hyperbola) as shown in the diagram since (in the same way as above) PV=constant, as pressure is increased volume decreases and vica versa. This inversely proportional relationship will (mathematically) always give a curve when plotted.</p><p>(If the maths is too much - learn the laws (Avogadro's, Boyle's, Charles' and Dalton's) on the revision slide.)</p><p>Thus<strong> P against V</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>How many moles of Iron will be produced if this reaction produces 500 mol of carbon dioxide?</p><p>Fe<sub>2</sub>O<sub>3</sub> + 3CO → 2Fe + 3CO<sub>2</sub></p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio"> 333 mol</label></p><p><label class="radio"><input type="radio"> 250 mol</label></p><p><label class="radio"><input type="radio"> 167 mol</label></p><p><label class="radio"><input type="radio"> 750 mol</label></p></div><div class="q-explanation"><p>The mole ratio must be used to answer this question:</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;">Fe<sub>2</sub>O<sub>3</sub></td><td style="text-align: center;">+</td><td style="text-align: center;">3CO</td><td style="text-align: center;">→</td><td style="text-align: center;">2Fe</td><td style="text-align: center;">+</td><td style="text-align: center;">3CO<sub>2</sub></td><td style="text-align: center;"> </td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">3</td><td style="text-align: center;">:</td><td style="text-align: center;"><strong>2</strong></td><td style="text-align: center;">:</td><td style="text-align: center;"><strong>3</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"> </td></tr></tbody></table><p>The reaction gives <strong>2 </strong>moles of iron (Fe) for every <strong>3</strong> moles of carbon dioxide (CO<sub>2</sub>) produced.</p><p>Therefore for every 1 mole of CO<sub>2</sub> produced, <span class="math-tex">\({2 \over 3}\)</span> of a mole of Fe is produced.</p><p>If 500 moles of CO<sub>2</sub> are produced, <span class="math-tex">\({2 \over 3}\)</span> × 500 = 333 moles of Fe will be produced.</p><p>Thus <strong>333 mol </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the sum of the smallest integer coefficients when ethene undergoes complete combustion?</p><p>___C<sub>2</sub>H<sub>4 </sub>+ ___O<sub>2 </sub>→ ___CO<sub>2</sub> + ___H<sub>2</sub>O</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>7</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>8</span></label> </p><p><label class="radio"> <input type="radio"> 6<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>9</span></label> </p></div><div class="q-explanation"><p>C<sub>2</sub>H<sub>4 </sub>+ 3O<sub>2 </sub>→ 2CO<sub>2</sub> + 2H<sub>2</sub>O is the balanced equation, so the sum of the integer coefficients is <strong>8</strong> - don't forget that C<sub>2</sub>H<sub>4</sub> does have a '1' in front of it, although by convention we don't actually write that into the equation.</p><p>(When balancing combustion reactions of hydrocarbons, first balance out carbon and hydrogen atoms by adding coefficients (numbers) in front of CO<sub>2 </sub>and H<sub>2</sub>O, after that balance the oxygen atoms and the hydrocarbon as required.)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What is the volume of an ideal gas when the pressure on 100cm<sup>3</sup> of gas is changed from 200kPa to 100kPa at constant temperature?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> 400 cm<sup>3</sup></label></p><p><label class="radio"><input class="c" type="radio"> 200 cm<sup>3</sup></label></p><p><label class="radio"><input type="radio"> 50 cm<sup>3</sup></label></p><p><label class="radio"><input type="radio"> 100 cm<sup>3</sup></label></p></div><div class="q-explanation"><p>The ideal gas equation is PV=nRT. The letters represent Pressure, Volume, number of moles, gas constant (R) and Temperature in that order.</p><p>If all other values in PV=nRT are constant then PV=constant. If we give that constant a value of 1 (PV=1) then P =<span class="math-tex">\({1 \over V}\)</span> so volume is inversely proportional to pressure (and P against <span class="math-tex">\({1\over V}\)</span> will give a straight line when plotted).</p><p>Therefore, if we halve the pressure, from 200 to 100 kPa, we must double the volume: Thus<strong> 200 cm<sup>3</sup> </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What is the concentration, in mol dm<sup>−3</sup>, of 10.0g of NaOH (M<sub>r</sub> = 40.0) in 500cm<sup>3</sup> of solution ?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>0.500 mol dm<sup>−3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>1.00 mol dm<sup>−3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>0.125 mol dm<sup>−3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>0.250 mol dm<sup>−3</sup></span></label> </p></div><div class="q-explanation"><p><strong>Concentration </strong>(mol dm<sup>−3</sup>) <strong>= moles </strong>(mol)<strong> / volume of solution </strong>(dm<sup>3</sup>)</p><p><strong>Moles </strong>(mol) = <strong>Mass </strong>(g) / <strong>molar mass </strong>(g mol<sup>−1</sup>)</p><p>So moles of sodium hydroxide (NaOH) = 10.0/40.0 = 0.250</p><p>Concentration of the solution = 0.250 / 0.500 (The volume must be converted to dm<sup>3</sup>: 1 dm<sup>3</sup> = 1000cm<sup>3</sup>)</p><p>Concentration = 0.250 / 0.500 = <strong>0.500 mol dm<sup>−3</sup></strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">11</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What volume of carbon dioxide can be obtained by reacting 500 cm<sup>3</sup> of methane (CH<sub>4</sub>) with 500 cm<sup>3 </sup>of oxygen (O<sub>2</sub>)?</p><p>CH<sub>4(g)</sub> + 2O<sub>2(g)</sub> → CO<sub>2(g)</sub> + 2H<sub>2</sub>O<sub>(l)</sub></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>500 cm<sup>3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>1000 cm<sup>3</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>2000 cm<sup>3</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>250 cm<sup>3</sup></span></label> </p></div><div class="q-explanation"><p>Ideal gases have a constant molar volume (at any given temperature and pressure).</p><p>Quantities of both reagents are given, which suggests that one reagent will be in excess and the other will be limiting. This is best considered in terms of the mole ratio:</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;">CH<sub>4</sub></td><td style="text-align: center;">+</td><td style="text-align: center;">2O<sub>2</sub></td><td style="text-align: center;">→</td><td style="text-align: center;">CO<sub>2</sub></td><td style="text-align: center;">+</td><td style="text-align: center;">2H<sub>2</sub>O</td><td> </td><td> </td></tr><tr><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">2</td><td style="text-align: center;">:</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">2</td><td> </td><td> </td></tr></tbody></table><p>1 mole of CH<sub>4</sub> will react with 2 moles of O<sub>2</sub>.</p><p>So methane will react with twice its volume of oxygen (and vice versa, oxygen will react with half its volume of methane).</p><p>Methane gas is therefore in excess and oxygen is the limiting reagent.</p><p>The amount of methane that will react is therefore 500 cm<sup>3</sup> / 2 = 250cm<sup>3</sup> (1:2 ratio)</p><p>And 250 cm<sup>3</sup> of methane reacting with give 250 cm<sup>3</sup> of carbon dioxide (CO<sub>2</sub>) (1:1 ratio)</p><p>Thus the answer is <strong>250cm<sup>3</sup></strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">12</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>How many moles of CaSO<sub>4</sub> are required to produce 128 g of SO<sub>2</sub>? (Ar: S = 32, O = 16)</p><p>3CaSO<sub>4</sub> + CaS → 4CaO + 4SO<sub>2</sub></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> 2.7 mol</label></p><p><label class="radio"><input class="c" type="radio"> 1.5 mol</label></p><p><label class="radio"><input type="radio"> 3.0 mol</label></p><p><label class="radio"><input type="radio"> 2.0 mol</label></p></div><div class="q-explanation"><p>The mole ratio must be used to answer this question:</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;">3CaSO<sub>4</sub></td><td style="text-align: center;">+</td><td style="text-align: center;">CaS</td><td style="text-align: center;">→</td><td style="text-align: center;">4CaO</td><td style="text-align: center;">+</td><td style="text-align: center;">4SO<sub>2</sub></td><td style="text-align: center;"> </td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;"><strong>3</strong></td><td style="text-align: center;">:</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">4</td><td style="text-align: center;">:</td><td style="text-align: center;"><strong>4</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"> </td></tr></tbody></table><p>The reaction produces <strong>4 </strong>moles of SO<sub>2</sub> for every <strong>3</strong> moles of CaSO<sub>4</sub> reacted.</p><p>Therefore for every 1 mole of SO<sub>2</sub> produced, <span class="math-tex">\({3 \over 4}\)</span> of a mole of CaSO<sub>4</sub> must be reacted.</p><p>Using <strong>Moles </strong>(mol) = <strong>mass</strong> (g) /<strong> molar mass </strong>(g mol<sup>−1</sup>) (molar mass of SO<sub>2</sub> = 32 + 16×2 = 64) 128g of SO<sub>2</sub> is 128 / 64 = 2.0 moles</p><p>If 2.0 moles of SO<sub>2</sub> is produced, <span class="math-tex">\({3 \over 4}\)</span> × 2.0 = 1.5 moles of CaSO<sub>4</sub> will be required.</p><p>Thus <strong>1.5 mol </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">13</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>24 g of bromine react with 18.1 g of metal, M, to form MBr<sub>3</sub>. What is the relative atomic mass of metal M? (A<sub>r</sub>: Br = 80)</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> 54.3</label></p><p><label class="radio"><input class="c" type="radio"> 181</label></p><p><label class="radio"><input type="radio"> 60.0</label></p><p><label class="radio"><input type="radio"> 90.5</label></p></div><div class="q-explanation"><p>The formula of the metal bromide is MBr<sub>3</sub>. This tells us that one mole of metal atoms will combine with three moles of bromine atoms (1:3).</p><p>Using <strong>Moles </strong>(mol) = <strong>mass </strong>(g) / <strong>molar mass </strong>(g mol<sup>−1</sup>), 24 g of bromine is 24 / 80 = 0.3 moles (the common multiple for mental arithmetic here is 8)</p><p>Using the ratio 3:1, 0.3 moles of bromine combines with 0.1 moles of metal, M.</p><p>So if 0.1 moles of metal M has a mass of 18.1 g (given in the question) then 1 mole of metal M will have a mass of 18.1 × 10 = 181 g</p><p>Therefore the relative atomic mass of metal M is <strong>181</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">14</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What is the molecular formula of a hydrocarbon containing 85.6% carbon by mass with an integer molar mass of 168g mol<sup>−1</sup>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>C<sub>13</sub>H<sub>12</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>6</sub>H<sub>12</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C<sub>12</sub>H<sub>24</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>7</sub>H<sub>14</sub></span></label> </p></div><div class="q-explanation"><p>We need to find the simplest ratio of the number of (moles of) atoms. We cannot use a calculator so we will need to approximate the relative atomic masses that we take from the periodic table (e.g. carbon 12.01 = 12).</p><p>Start by assuming 100g: Thus 85.6g of C atoms, 14.4g of H atoms:</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Carbon</td><td style="text-align: center;">Hydrogen</td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;"><p>moles = mass/molar mass</p></td><td style="text-align: center;">85.6/12 = A little over 7 (7×12=84)</td><td style="text-align: center;">14.4/1 = 14.4</td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;">Approximate mole ratio</td><td style="text-align: center;">7...</td><td style="text-align: center;">14...</td><td style="text-align: center;"> </td></tr><tr><td style="text-align: center;">Simplest mole ratio (empirical formula)</td><td style="text-align: center;"><strong>1</strong></td><td style="text-align: center;"><strong>2</strong></td><td style="text-align: center;"> </td></tr></tbody></table><p>This suggests that the empirical formula is probably CH<sub>2</sub>. The empirical formula mass is therefore approximately 14 units (12 + 1+ 1).</p><p>168g mol<sup>−1</sup> is the molar mass (rounded to a whole number) and 168 / 14 = 12</p><p>So there are 12 empirical formula units (of CH<sub>2</sub>) in the molecular formula.</p><p>Therefore the molecular formula is <strong>C<sub>12</sub>H<sub>24</sub></strong>.</p><p>(Once the approximate 1:2 ratio of C:H is known, the last part can also be done by trial and error: 12×12 = 144, 144+24 = 168)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">15</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What is the percentage yield (%) when 14 g of ethene produces 12 g of ethanol?</p><p>C<sub>2</sub>H<sub>4</sub> + H<sub>2</sub>O → C<sub>2</sub>H<sub>5</sub>OH</p><p>Using Mr(ethene) = 28 and Mr(ethanol) = 46.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({12\times46\times100 \over 14\times28}\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\({12\times28\times100 \over 14\times46}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({14\times12\times100 \over 28\times46}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({14\times28\times100 \over 12\times46}\)</span></span></label> </p></div><div class="q-explanation"><p><strong>Percentage yield</strong> = (experimental yield / theoretical yield) × 100%</p><p><strong>Moles</strong> = <strong>mass </strong>/ <strong>molar mass</strong></p><p>Moles of ethene = 14 / 28</p><p>Moles of ethanol = 12 / 46</p><p>The reaction has a 1:1 molar ratio as seen in the equation. So the theoretical yield of ethanol in moles will be the same as the moles of ethene.</p><p>Therefore the percentage yield will be: (moles of ethanol / moles of ethene) × 100%</p><p>= (12 / 46) / (14 / 28) × 100%</p><p>(The easiest way to deal with this is probably to remember that dividing by a fraction is the same as multiplying by the inverted (upside down) fraction. So...)</p><p>= <span class="math-tex">\({12\over 46}\)</span> × <span class="math-tex">\({28\over 14}\)</span> × 100%</p><p>= <span class="math-tex">\({12\times28\times100 \over 14\times46}\)</span><strong> </strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">16</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator.</em></p><p>What is the value of x when 36.0g of MnCl<sub>2</sub>.xH<sub>2</sub>O is heated to dryness leaving 31.5g of anhydrous MnCl<sub>2</sub>?</p><p>Using A<sub>r</sub>: Mn = 55, Cl = 35.5, H = 1, O = 16</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> 0.25</label></p><p><label class="radio"><input type="radio"> 10</label></p><p><label class="radio"><input type="radio"> 4</label></p><p><label class="radio"><input class="c" type="radio"> 1</label></p></div><div class="q-explanation"><p>The value for the water of crystallisation, x, can be found by finding the integer ratio of <strong>moles of MnCl<sub>2</sub> : moles of H<sub>2</sub>O</strong>. (x is always given as a whole number.)</p><p><strong>Moles</strong> (mol) = <strong>mass </strong>(g) / <strong>molar mass </strong>(g mol<sup>−1</sup>)</p><p>The moles of MnCl<sub>2</sub> can be found using the mass of the anhydrous salt: 31.5 / 126 = <span class="math-tex">\({1\over 4}\)</span> = 0.25 mol</p><p>The mass of water can be found by subtracting the mass of anhydrous salt from hydrated salt: 36.0 − 31.5 = 4.5 g</p><p>And moles of water = 4.5 / 18 = <span class="math-tex">\({1\over 4}\)</span> = 0.25 mol</p><p>The ratio of <strong>moles of MnCl<sub>2</sub> : moles of H<sub>2</sub>O</strong> is therefore 0.25 : 0.25 and the integer ratio is therefore 1 : 1.</p><p>The value of x is therefore <strong>1</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> <div class="page-container panel-self-assessment" data-id="2123"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>Stoichiometry core (SL & HL) paper 1 questions</strong> have you understood?</p><div class="slider-container text-center"><div id="self-assessment-slider" class="sib-slider self-assessment " data-value="1" data-percentage=""></div></div> <label class="label-lg">My notes</label> <textarea name="page-notes" class="form-control" rows="3" placeholder="Write your notes here..."></textarea> </div> <div class="panel-footer text-xs-center"> <span id="last-edited" class="mb-xs-3"> </span> <div class="actions mt-xs-3"> <button id="save-my-progress" type="button" class="btn btn-sm btn-primary text-center btn-xs-block"> <i class="fa fa-fw fa-floppy-o"></i> Save </button> </div> </div></div> <div id="modal-feedback" class="modal fade" tabindex="-1" role="dialog"> <div class="modal-dialog" role="document"> <div class="modal-content"> <div class="modal-header"> <h4 class="modal-title">Feedback</h4> <button type="button" class="close hidden-xs hidden-sm" data-dismiss="modal" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div> <div class="modal-body"> <div class="errors"></div> <p><strong>Which of the following best describes your feedback?</strong></p> <form method="post" style="overflow: hidden"> <div class="form-group"> <div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Recommendation"> Recommend</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Problem"> Report a problem</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Improvement"> Suggest an improvement</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Other"> Other</label></div> </div> <hr> <div class="row"> <div class="col-md-6"> <div class="form-group"> <label for="feedback-name">Name</label> <input type="text" class="form-control" name="feedback-name" placeholder="Name" value=" "> </div> </div> <div class="col-md-6"> <div class="form-group"> <label for="feedback-email">Email address</label> <input type="email" class="form-control" name="feedback-email" placeholder="Email" value="@airmail.cc"> </div> </div> </div> <div class="form-group"> <label for="feedback-comments">Comments</label> <textarea class="form-control" name="feedback-comments" style="resize: vertical;"></textarea> </div> <input type="hidden" name="feedback-ticket" value="082b9c9c4ae3624d"> <input type="hidden" name="feedback-url" value="https://studyib.net/chemistry/page/2123/stoichiometry-core-sl-hl-paper-1-questions"> <input type="hidden" name="feedback-subject" value="7"> <input type="hidden" name="feedback-subject-name" value="Chemistry"> <div class="pull-left"> </div> </form> </div> <div class="modal-footer"> <button type="button" class="btn btn-primary btn-xs-block feedback-submit mb-xs-3 pull-right"> <i class="fa fa-send"></i> Send </button> <button type="button" class="btn btn-default btn-xs-block m-xs-0 pull-left" data-dismiss="modal"> Close </button> </div> </div> </div></div> <style type="text/css" media="screen">/* Important part */
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