A 6.53 -g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces \(1.72 \mathrm{~L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and 743 torr pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

For the reaction of hydrochloric acid (HCl) with magnesium carbonate (MgCO3): \[MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + CO_2(g) + H_2O(l)\] For the reaction of hydrochloric acid (HCl) with calcium carbonate (CaCO3): \[CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)\]

We can use the Ideal Gas Law formula, which is: \(PV = nRT\) Where: - P = pressure (in atm) - V = volume (in L) - n = moles (in mol) - R = ideal gas constant (0.0821 L⋅atm/mol⋅K) - T = temperature (in K) First, we convert the given pressure from torr to atm: \[743 \,\text{torr} \times \frac{1 \,\text{atm}}{760\, \text{torr}} = 0.976 \,\text{atm}\] Next, we convert the given temperature from Celsius to Kelvin: \[(28^\circ \text{C}) + 273.15 = 301.15 \,\text{K}\] Now, we can use the Ideal Gas Law formula to find the total moles (n) of carbon dioxide (CO2) formed: \[n_\text{CO2} = \frac{PV}{RT} = \frac{(0.976 \,\text{atm})(1.72\, \text{L})}{(0.0821 \, \text{L⋅atm/mol⋅K})(301.15 \,\text{K})} = 0.0732\, \text{mol}\]

Let x be the mass of magnesium carbonate (MgCO3) and (6.53 - x) be the mass of calcium carbonate (CaCO3) in the mixture. Then, the moles of CO2 produced from each component can be calculated as: For MgCO3: \(n_\text{CO2(MgCO3)}=\frac{x}{M_\text{MgCO3}}\) For CaCO3: \(n_\text{CO2(CaCO3)}=\frac{6.53-x}{M_\text{CaCO3}}\) The total moles of CO2 formed from both components is 0.0732 mol (from part b), so: \(\frac{x}{M_\text{MgCO3}} + \frac{6.53 - x}{M_\text{CaCO3}} = 0.0732\) Where: - \(M_\text{MgCO3}\) is the molar mass of magnesium carbonate, which is 84.31 g/mol. - \(M_\text{CaCO3}\) is the molar mass of calcium carbonate, which is 100.09 g/mol. Solve the equation for x: \(x = 3.67 \,\text{g}\) Finally, we can calculate the percentage by mass of magnesium carbonate in the mixture: \[\text{Percentage of MgCO3} = \frac{3.67 \,\text{g}}{6.53 \,\text{g}} \times 100\% = 56.2\%\]