Step 1

$\mathbf{PV}\mathbf{}\mathbf{=}\mathbf{}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{PV}}{\mathbf{RT}}$

$\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{(744\mathrm{torr}\times {\displaystyle \frac{1\mathrm{atm}}{760\mathrm{torr}}})(1.73L)}{(0.08206{\displaystyle \frac{\mathrm{atm}\xb7L}{\mathrm{mol}\xb7K}})(28\xb0C+273.15)\mathbf{K}}$

**n = 0.0685 mol CO _{2}**

A 6.54 g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces 1.73 L of carbon dioxide gas at 28 ^{o}C and 744 torr pressure.

Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

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