17. When a 16.8 -gram sample of an unknown mineral was dissolved in acid, 4.4 -grams of $$ \mathrm{CO}_{2} $$ were generated. If the rock contained no carbonate other than $$ \mathrm{MgCO}_{3} $$, what was the percent of MgCO by mass in the limestone? Molar mass $$ (\mathrm{g} / \mathrm{mol}): \mathrm{MgCO}_{3}=84 $$ and $$ \mathrm{CO}_{2}=44 $$

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1. Write down the reaction information. Since the only carbonate in the limestone is $$ \mathrm{MgCO}_{3} $$, the reaction produces $$ \mathrm{CO}_{2} $$ with a 1:1 molar ratio:

$$
   \mathrm{MgCO}_{3} \rightarrow \mathrm{CO}_{2} + \text{other products}
   $$

2. Calculate the moles of $$ \mathrm{CO}_{2} $$ generated:

$$
   \text{Moles of } \mathrm{CO}_{2} = \frac{4.4\ \text{g}}{44\ \text{g/mol}} = 0.1\ \text{mol}
   $$

3. Since the molar ratio of $$ \mathrm{MgCO}_{3} $$ to $$ \mathrm{CO}_{2} $$ is 1:1, the moles of $$ \mathrm{MgCO}_{3} $$ present in the sample is also $$ 0.1 $$ moles.

4. Calculate the mass of $$ \mathrm{MgCO}_{3} $$:

$$
   \text{Mass of } \mathrm{MgCO}_{3} = 0.1\ \text{mol} \times 84\ \text{g/mol} = 8.4\ \text{g}
   $$

5. Calculate the percent of $$ \mathrm{MgCO}_{3} $$ by mass in the limestone:

$$
   \text{Percent} = \left(\frac{8.4\ \text{g}}{16.8\ \text{g}}\right) \times 100\% = 50\%
   $$

Thus, the percentage of $$ \mathrm{MgCO}_{3} $$ in the limestone is $$ 50\% $$.
The limestone contains 50% $$ \mathrm{MgCO}_{3} $$ by mass.

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