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For the reaction between hydrochloric acid (HCl) and magnesium carbonate (MgCO3), the balanced chemical equation is given by: \[ \mathrm{MgCO_3(s) + 2HCl (aq) \longrightarrow MgCl_2(aq) + H_2O(l) + CO_2(g)} \] For the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3), the balanced chemical equation is given by: \[ \mathrm{CaCO_3(s) + 2HCl (aq) \longrightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)} \]

Given the volume of CO2 produced is 1.72 L, temperature is 28°C (convert to Kelvin: 28 + 273.15 = 301.15 K), and pressure is 743 torr (convert to atm: 743 ÷ 760 = 0.9776 atm), we can use the ideal gas law formula: \[PV=nRT\] Where: P = pressure in atm V = volume in L n = number of moles R = ideal gas constant (\(0.0821\frac{\mathrm{atm·L}}{\mathrm{K·mol}}\)) T = temperature in K Rearrange the formula to find n: \[n=\frac{PV}{RT}\] Plugging in the values, we get: \[n=\frac{(0.9776 \mathrm{~atm})(1.72 \mathrm{~L})}{(0.0821 \frac{\mathrm{atm·L}}{\mathrm{K·mol}})(301.15 \mathrm{~K})} \] \[n\approx 0.0725 \mathrm{~mol}\] There are approximately 0.0725 moles of CO2 produced from the reaction.

Let x be the moles of magnesium carbonate (MgCO3) in the mixture and y be the moles of calcium carbonate (CaCO3) in the mixture. Since the total number of CO2 moles produced from both reactions is 0.0725, we have: \[x + y = 0.0725\] We also know the mass of the entire mixture is 6.53 g. Using the molar mass of MgCO3 (approximately 84.3 g/mol) and CaCO3 (approximately 100.1 g/mol), we can write the equation: \[84.3x + 100.1y = 6.53\] Now, we need one more equation to solve for x and y. We use the ratio of magnesium to carbon dioxide from the balanced equations: \[x = 0.5 \times 0.0725 = 0.03625\] Now, we can substitute x into the first equation: \[0.03625 + y = 0.0725\] \[y \approx 0.03625\] Now that we have the moles of each carbonate, we can calculate the mass of magnesium carbonate and find the percentage: Mass of MgCO3 = (0.03625 mol) × (84.3 g/mol) ≈ 3.056 g Percentage of MgCO3 in mixture = \(\frac{3.056 \mathrm{~g}}{6.53 \mathrm{~g}} \times 100 \approx 46.8 \% \) The percentage by mass of magnesium carbonate in the mixture is approximately 46.8%.