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 \( \mathrm{CO}_{2}=44 \)

To find the percent of \( \text{MgCO}_3 \) by mass in the limestone, we first need to determine the moles of \( \text{CO}_2 \) generated from the reaction. Given the mass of \( \text{CO}_2 \) (4.4 g) and its molar mass (44 g/mol), we can calculate the number of moles: \[ \text{Moles of } \text{CO}_2 = \frac{4.4 \text{ g}}{44 \text{ g/mol}} = 0.1 \text{ mol} \] Since the reaction shows that one mole of \( \text{MgCO}_3 \) produces one mole of \( \text{CO}_2 \), we conclude that 0.1 moles of \( \text{MgCO}_3 \) were present in the sample. Next, we calculate the mass of \( \text{MgCO}_3 \): \[ \text{Mass of } \text{MgCO}_3 = 0.1 \text{ mol} \times 84 \text{ g/mol} = 8.4 \text{ g} \] Finally, we find the percent by mass of \( \text{MgCO}_3 \) in the limestone: \[ \text{Percent of } \text{MgCO}_3 = \left(\frac{8.4 \text{ g}}{16.8 \text{ g}}\right) \times 100\% = 50\% \] So, the percent of \( \text{MgCO}_3 \) by mass in the limestone is 50%.