IB Chemistry SL R2.1 How much? The amount of chemical change questions

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Question 1

Multiple choiceHard1 mark

Magnesium reacts with hydrochloric acid according to the equation below. The hydrogen is collected at the conditions for which the molar gas volume is 24.0 dm3 mol−124.0\ \mathrm{dm^3\,mol^{-1}} Use Ar(Mg)=24.31A_\mathrm{r}(\mathrm{Mg})=24.31

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)}
SubstanceAmount
Mg0.120 g0.120\ \mathrm{g}
HCl(aq)25.0 cm325.0\ \mathrm{cm^3} of 0.200 mol dm−30.200\ \mathrm{mol\,dm^{-3}}
H2H_2 collected56.4 cm356.4\ \mathrm{cm^3}

Which option gives the limiting reactant and the percentage yield of H2H_2

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Answer: C. Compare the available amounts using the 1:21:2 mole ratio of Mg to HCl, then calculate the yield relative to the theoretical H2H_2 volume.

  1. n(Mg)=0.120/24.31=4.94×10−3 moln(\mathrm{Mg})=0.120/24.31=4.94\times10^{-3}\ \mathrm{mol} n(HCl)=0.0250×0.200=5.00×10−3 moln(\mathrm{HCl})=0.0250\times0.200=5.00\times10^{-3}\ \mathrm{mol} The Mg would require 9.87×10−3 mol9.87\times10^{-3}\ \mathrm{mol} HCl, so HCl is limiting.
  2. n(H2)=5.00×10−3/2=2.50×10−3 moln(\mathrm{H_2})=5.00\times10^{-3}/2=2.50\times10^{-3}\ \mathrm{mol} giving a theoretical volume of 60.0 cm360.0\ \mathrm{cm^3}
  3. Percentage yield =(56.4/60.0)×100=94.0%=(56.4/60.0)\times100=94.0\%

A: Compares the raw mole amounts to label Mg limiting, ignoring the equation’s coefficients.
B: Reverses the yield ratio, calculating 60.0/56.4×100=106%60.0/56.4\times100=106\%
D: Treats Mg as limiting and uses its amount to calculate the theoretical H2H_2 volume, giving 118.5 cm3118.5\ \mathrm{cm^3} and a 47.6%47.6\% yield.
Remember: compare available amounts after accounting for stoichiometric coefficients to identify the limiting reactant.

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