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class="mark-page-favorite pull-right" data-pid="724" 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="../343/stoichiometry.html">Stoichiometry</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Reacting masses and volumes</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 60 minutes"><i class="fa fa-clock-o"></i> 60'</span> </ol> <article id="main-article"> <p><img alt="" src="../../images/test-images/nahco3reaction-1.jpg" style="float: left; width: 160px; height: 107px;">There is not a great deal of knowledge to learn in order to be confident with mole calculations, but you may well need lots of practice. There are only certain ways in which questions on stoichiometry are asked in exams, so <strong>attempt as many questions as you can until you start to recognise the different types and feel more confident answering them</strong>.</p> <p>Remember that <em>error carried forward </em>is applied to when calculations are marked, so even if you make a mistake early in a question you can still score all the marks beyond!<strong> Use the table method in the revision cards to help you to organise your thoughts and your working.</strong></p> <hr class="hidden-separator"> <div class="panel panel-has-colored-body panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <p>Ensure you are confident using the terms below and learn the asterisked* definitions</p> <p>limiting reactant, excess reactant, theoretical yield, experimental yield, percentage yield, Avogadro's law*, ideal gas model, concentration, solution, standard solution</p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="" data-topics="590" data-subject-id="7" data-n-flashcards="10" style="text-align:center">Show flashcards</a></div><hr> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Mass-mass calculations</p> </div> </div> <div class="panel-body"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/378282682"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Calculating percentage yield</p> </div> </div> <div class="panel-body"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/378764155"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>The ideal gas equation: PV=nRT</p> </div> </div> <div class="panel-body"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/378283270"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Reacting gas volumes</p> </div> </div> <div class="panel-body"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/378506423"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> 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class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following are reasons why <em>real gases </em>deviate from the behaviour of <em>ideal gases</em> at, for example, very high pressures?</p><p><strong>1: </strong>Real gases do not fill a container uniformly.</p><p><strong>2: </strong>Real gases have intermolecular forces between the particles.</p><p><strong>3: </strong>Real gases have volume associated with the particles.</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p></div><div class="q-explanation"><p>Ideal gases are assumed to:</p><ul><li>be point particles with no volume;</li><li>move in straight lines with random motion;</li><li>behave as rigid spheres;</li><li>engage in perfectly elastic collisions;</li><li>have negligible forces between molecules.</li></ul><p>A further assumption is that temperature represents a measure of the average kinetic energy of particles in a system.</p><p>Of the reasons given:</p><p>Real gases <strong>do</strong> fill a container uniformly, so reason 1 is not correct.</p><p>Real gases do have intermolecular forces between the particles and do have volume, so reasons 2 and 3 are correct in that they contribute to why real gases do not behave ideally at high pressures.</p><p>Thus 2 and 3 only 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>5.00g of hex-1-ene was reacted with excess bromine at room temperature.</p><p>C<sub>6</sub>H<sub>12</sub> + Br<sub>2</sub> → C<sub>6</sub>H<sub>12</sub>Br<sub>2</sub></p><p>Giving the answer to two decimal places, calculate what mass (g) of 1,2-dibromohexane should be formed?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>9.49</span></label> </p><p><label class="radio"> <input type="radio"> <span>5.00</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>14.49</span></label> </p><p><label class="radio"> <input type="radio"> <span>14.51</span></label> </p></div><div class="q-explanation"> <p>Molar mass of C<sub>6</sub>H<sub>12</sub> = (12.01×6)+(1.01×12) = 84.18</p><p>Molar mass of C<sub>6</sub>H<sub>12</sub>Br<sub>2 </sub>= 84.18 (above) + 79.90×2 = 243.92</p><p>Moles of hex-1-ene = mass / molar mass = 5.00 / 84.18 = 0.059396531</p><p>1:1 ratio of C<sub>6</sub>H<sub>12 </sub>to C<sub>6</sub>H<sub>12</sub>Br<sub>2</sub>, so moles of C<sub>6</sub>H<sub>12</sub>Br<sub>2 </sub>= 0.059396531</p><p>Mass of 1,2-dibromohexane = moles × molar mass = 0.059396531 × 243.92 = 14.4880019</p><p>Thus the answer is 14.49 (2 dp)</p><p><strong>Incorrect answers:</strong></p><p>14.51 can be obtained by using 12 and 1 as the molar masses of carbon and hydrogen respectively.</p><p>9.49 is the mass of bromine that will react with the hexene.</p><p>5.00 is the starting mass of the hexene re-stated.</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>Hydrogen and oxygen gases react together to produce water:</p><p>2H<sub>2(g)</sub> + O<sub>2(g)</sub> → 2H<sub>2</sub>O<sub>(l)</sub></p><p>If 600cm<sup>3</sup> of hydrogen and 500cm<sup>3</sup> of oxygen are reacted to produce as much liquid water as possible, which gas/es will be left over at the end?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>100cm<sup>3</sup> of hydrogen</span></label> </p><p><label class="radio"> <input type="radio"> <span>400cm<sup>3</sup> of hydrogen and 400cm<sup>3</sup> of oxygen</span></label> </p><p><label class="radio"> <input type="radio"> <span>None</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>200cm<sup>3</sup> of oxygen</span></label> </p></div><div class="q-explanation"><p>Ideal gases have a constant molar volume (at any given temperature and pressure).</p><p>The mole ratio here is 2H<sub>2</sub> : 1O<sub>2</sub>. So oxygen will react with twice its volume of hydrogen (and vice versa, hydrogen will react with half its volume of oxygen).</p><p>To give the maximum amount of water, all 600cm<sup>3</sup> of hydrogen will react with 300cm<sup>3 </sup>of oxygen. Oxygen gas is therefore in excess.</p><p>The amount of oxygen left over will be 500 − 300 = 200cm<sup>3</sup></p><p>Thus the answer is 200cm<sup>3</sup> of oxygen</p><p><strong>Incorrect answers:</strong></p><p>100cm<sup>3</sup> hydrogen, assumes reacting in a 1:1 ratio (rather than 2:1).</p><p>400cm<sup>3</sup> of hydrogen and 400cm<sup>3</sup> of oxygen, suggests reacting in the correct ratio, but not producing a maximum amount of water.</p><p>None, assumes that all of the gases will react, which is incorrect.</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>A student carries out an experiment to make cyclohexene (C<sub>6</sub>H<sub>10</sub>) from cyclohexanol (C<sub>6</sub>H<sub>12</sub>O).</p><p>The student uses 6.50g of cyclohexanol and obtains 3.15g of cyclohexene.</p><p>What is the percentage yield of the experiment (in %)?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>59</span></label> </p><p><label class="radio"> <input type="radio"> <span>5.33</span></label> </p><p><label class="radio"> <input type="radio"> <span>40</span></label> </p><p><label class="radio"> <input type="radio"> <span>169</span></label> </p></div><div class="q-explanation"><p>Molar masses:</p><p>C<sub>6</sub>H<sub>10</sub> is (12.01×6)+(1.01×10) = 82.16</p><p>C<sub>6</sub>H<sub>12</sub>O is 82.16 (above) + (1.01×2)+16.00 = 100.18</p><p>The reaction has a 1:1 molar ratio.</p><p>moles of cyclohexanol = 6.50 / 100.18 = 0.06488321</p><p><strong>theoretical yield</strong> of cyclohexene = moles × molar mass = 0.06488321 × 82.16 = 5.330804552g</p><p><strong>experimental yield</strong> is given: 3.15g</p><p><strong>percentage yield</strong> = (experimental yield / theorectical yield) × 100%</p><p>= (3.15 / 5.330804552) × 100% = 59.0905175%</p><p><strong>Incorrect answers</strong></p><p>5.33 is simply stating the theoretical yield (in g).</p><p>169 is assuming the percentage yield equation is inverted (5.33.../3.15).</p><p>40 assumes the molar masses are swapped around in the calculation (82.16 used instead of 100.18 and vice versa).</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>2.90g of a volatile liquid was evaporated and found to occupy approximately 1217cm<sup>3</sup> at 20°C and atmospheric pressure (100kPa). To two significant figures, what is the molar mass of the volatile liquid?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>4.0</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>58</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.0040</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.058</span></label> </p></div><div class="q-explanation"><p>Use PV=nRT (Gas constant, R, given in the data book as 8.31 J K<sup>−1</sup> mol<sup>−1</sup>)</p><p>Units are critical; pressure in kPa and volume in dm<sup>3</sup> work correctly (or Pa and m<sup>3</sup>) and temperature must be in Kelvin (°C + 273).</p><p>PV=nRT</p><p>100 × 1.217 = n × 8.31 × 293</p><p>n = 100 × 1.217 / 8.31 × 293 = 121.7 / 2434.83 = 0.05 mol (approximately)</p><p>molar mass = mass / mol</p><p>molar mass = 2.90 / 0.05 = 58 g mol<sup>−1</sup></p><p>Thus 58 is the corect answer</p><p><strong>Incorrect answers</strong></p><p>4 assumes a temperature of 20 (rather than 293).</p><p>0.058 assumes a volume of 1217 (rather than 1.217).</p><p>0.0040 assumes both a temperature of 20 (rather than 293) and a volume of 1217 (rather than 1.217).</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>Lead nitrate solution and potassium iodide solution react when mixed to precipitate out lead iodide, a yellow solid, which can be filtered off:</p><p>Pb(NO<sub>3</sub>)<sub>2(aq)</sub> + 2KI<sub>(aq)</sub> → PbI<sub>2(s)</sub> + 2KNO<sub>3(aq)</sub></p><p>If 200cm<sup>3</sup> of 0.100 mol dm<sup>−3</sup> lead nitrate is added to 500cm<sup>3</sup> of 0.200 mol dm<sup>−3</sup> potassium iodide, what mass (in g) of lead iodide will be produced?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>9.22</span></label> </p><p><label class="radio"> <input type="radio"> <span>46.10</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.0200</span></label> </p><p><label class="radio"> <input type="radio"> <span>6.68</span></label> </p></div><div class="q-explanation"><p>The quantities of both reactants are given, so this suggests that one reactant will be in excess.</p><p>The reaction has a 1:2 molar ratio of reactants.</p><p>moles = concentration × volume (dm<sup>3</sup>)</p><p>moles of lead nitrate = 0.100 × 0.200 = 0.0200mol</p><p>moles of potassium iodide = 0.200 × 0.500 = 0.100mol</p><p>potassium iodide is therefore in excess (×5 amount of lead nitrate, when only ×2 is needed to react completely).</p><p>All of the 0.0200mol of lead nitrate will react to produce 0.0200mol of lead iodide (1:1 ratio).</p><p>molar mass of lead iodide = 207.20 + (2×126.90) = 461.00</p><p>mass of lead iodide = moles × molar mass = 0.0200 × 461.00 = 9.22g</p><p>Thus the correct answer is 9.22</p><p><strong>Incorrect answers</strong></p><p>46.10 assumes 0.100mol of lead iodide is produced (rather than 0.0200).</p><p>6.68 assumes the molar mass of lead iodide uses a formula of PbI (rather than PbI<sub>2</sub>).</p><p>0.0200 states the moles of lead iodide (rather than the mass).</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>A titration can be used to measure the concentration of oxalic acid (ethandioic acid) in aqueous solution according to the following equation. Phenolphthalein is used as an indicator.</p><p>(COOH)<sub>2(aq)</sub> + 2NaOH<sub>(aq) </sub>→ (COONa)<sub>2(aq)</sub> + 2H<sub>2</sub>O<sub>(l)</sub></p><p>If an average of 21.50cm<sup>3</sup> of 1.00 mol dm<sup>−3</sup> sodium hydroxide is needed to neutralise 12.50cm<sup>3</sup> of oxalic acid solution, what is the concentration of the oxalic acid? Give your answer in <strong>g dm<sup>−3</sup></strong>.</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>77.4</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.86</span></label> </p><p><label class="radio"> <input type="radio"> 309.7<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>38.7</span></label> </p></div><div class="q-explanation"><p>The reaction has a 1:2 molar ratio of reactants.</p><p>moles = concentration × volume (dm<sup>3</sup>)</p><p>moles of sodium hydroxide = 1.00 × 0.02150 = 0.02150mol</p><p>moles of oxalic acid = 0.02150 ÷ 2 = 0.01075mol</p><p>concentration of oxalic acid = moles / volume (dm<sup>3</sup>)</p><p>= 0.01075 / 0.01250 = 0.8600 mol dm<sup>−3</sup></p><p>molar mass of oxalic acid = (12.01 + 16.00 + 16.00 + 1.01) × 2 = 90.04</p><p>concentration of oxalic acid in g dm<sup>−3</sup> = concentration in mol dm<sup>−3</sup> × molar mass = 0.8600 × 90.04</p><p>= 77.4344 g dm<sup>−3</sup></p><p>Thus the correct answer is 77.4</p><p><strong>Incorrect answers</strong></p><p>309.7 assumes moles of sodium hydroxide is ×2 instead of ÷2 (mole ratio the wrong way around).</p><p>38.7 assumes the molar mass of oxalic acid is <strong>half </strong>its actual value (the 2 in (COOH)<sub>2</sub> has been ignored).</p><p>0.86 states the concentraion in mol dm<sup>−3</sup> (rather than in g dm<sup>−3</sup>).</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> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a 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