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Magnesium carbonate reacts with hydrochloric acid to form magnesium chloride, carbon dioxide, and water. The balanced chemical equation for this reaction is: \[MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + CO_2(g) + H_2O(l)\] Similarly, calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water. The balanced chemical equation for this reaction is: \[CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)\]

We are given the volume, temperature, and pressure of the carbon dioxide gas produced. We can use the ideal gas law equation to calculate the moles of CO2: \(PV=nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. The gas constant R for this equation is 0.0821 L atm / (mol K). We'll convert the given pressure from torr to atm (as 1 atm = 760 torr) and the temperature from Celsius to Kelvin (by adding 273.15). Pressure in atm = \(743 \,\mathrm{torr} \times \frac{1 \,\mathrm{atm}}{760 \,\mathrm{torr}} = 0.9776 \,\mathrm{atm}\) Temperature in K = \(28^{\circ} \mathrm{C} + 273.15 = 301.15 \,\mathrm{K}\) Now, we can plug in the values into the ideal gas law equation and solve for n. \(0.9776\,\mathrm{atm} \times 1.72\,\mathrm{L} = n \times 0.0821 \,\frac{\mathrm{L\,atm}}{\mathrm{mol\,K}} \times 301.15\, \mathrm{K}\) Solve for n: \[n=\frac{0.9776 \,\mathrm{atm}\times1.72\,\mathrm{L}}{0.0821\,\frac{\mathrm{L\, atm}}{\mathrm{mol\, K}}\times301.15\,\mathrm{K}} = 0.0723 \,\mathrm{moles}\] So, 0.0723 moles of carbon dioxide are produced.

Let x represent the moles of magnesium carbonate in the mixture. Then, the moles of calcium carbonate in the mixture will be 0.0723 - x. Using the balanced equations from step 1, we can write a mass equation for the mixture : \(6.53\,\mathrm{g} = x\times M_{MgCO_3} + (0.0723-x)\times M_{CaCO_3}\) Where \(M_{MgCO_3}\) and \(M_{CaCO_3}\) are the molar masses of magnesium carbonate and calcium carbonate, respectively. \(M_{MgCO_3} = 24.3\,\mathrm{g/mol} + 12.01\,\mathrm{g/mol} + 3 \times 16.00\,\mathrm{g/mol} = 84.31\,\mathrm{g/mol}\) \(M_{CaCO_3} = 40.08\,\mathrm{g/mol} + 12.01\,\mathrm{g/mol} + 3 \times 16.00\,\mathrm{g/mol} = 100.09\,\mathrm{g/mol}\) Now, we can plug in the molar masses into the mass equation and solve for x: \(6.53\,\mathrm{g} = x\times 84.31\,\mathrm{g/mol} + (0.0723-x)\times 100.09\,\mathrm{g/mol}\) Solve for x: \[x=\frac{6.53\: \mathrm{g} -0.0723\times100.09\: \mathrm{g/mol}}{84.31\:\mathrm{g/mol}-100.09\:\mathrm{g/mol}}= 0.0296\:\mathrm{moles}\] Now, we can determine the mass of magnesium carbonate in the mixture: Mass of \(MgCO_{3} = 0.0296\,\mathrm{moles} \times 84.31\,\mathrm{g/mol} = 2.49\,\mathrm{g}\) Finally, we can calculate the percentage by mass of magnesium carbonate in the mixture: Percentage of \(MgCO_{3} = \frac{2.49\,\mathrm{g}}{6.53\,\mathrm{g}} \times 100\% = 38.13\% \) Thus, the percentage of magnesium carbonate in the mixture is 38.13%.