IB Chemistry SL R2.1 practice question

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 $24.0\ \mathrm{dm^3\,mol^{-1}}$ Use $A_\mathrm{r}(\mathrm{Mg})=24.31$

\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)}

Substance	Amount
Mg	$0.120\ \mathrm{g}$
HCl(aq)	$25.0\ \mathrm{cm^3}$ of $0.200\ \mathrm{mol\,dm^{-3}}$
$H_2$ collected	$56.4\ \mathrm{cm^3}$

Which option gives the limiting reactant and the percentage yield of $H_2$

Show answer and explanation

Answer: C. Compare the available amounts using the $1:2$ mole ratio of Mg to HCl, then calculate the yield relative to the theoretical $H_2$ volume.

$n(\mathrm{Mg})=0.120/24.31=4.94\times10^{-3}\ \mathrm{mol}$ $n(\mathrm{HCl})=0.0250\times0.200=5.00\times10^{-3}\ \mathrm{mol}$ The Mg would require $9.87\times10^{-3}\ \mathrm{mol}$ HCl, so HCl is limiting.
$n(\mathrm{H_2})=5.00\times10^{-3}/2=2.50\times10^{-3}\ \mathrm{mol}$ giving a theoretical volume of $60.0\ \mathrm{cm^3}$
Percentage yield $=(56.4/60.0)\times100=94.0\%$

A: Compares the raw mole amounts to label Mg limiting, ignoring the equation’s coefficients.
B: Reverses the yield ratio, calculating $60.0/56.4\times100=106\%$
D: Treats Mg as limiting and uses its amount to calculate the theoretical $H_2$ volume, giving $118.5\ \mathrm{cm^3}$ and a $47.6\%$ yield.
Remember: compare available amounts after accounting for stoichiometric coefficients to identify the limiting reactant.