Problem 128 - Chemistry: The Central Science

Question

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.

Step-by-Step Solution

Answer

The balanced chemical equations for the reactions between hydrochloric acid and magnesium carbonate, and hydrochloric acid and calcium carbonate are: \[MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + CO_2(g) + H_2O(l)\] \[CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)\] Using the ideal gas law, we calculate the total number of moles of carbon dioxide produced to be 0.0723 moles. Upon determining the percentage by mass of magnesium carbonate in the mixture, we find it to be approximately 38.13%.

Step 1: Write balanced chemical equations for the reactions of hydrochloric acid with magnesium carbonate and calcium carbonate

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: \[CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)\]

Step 2: Calculate the total number of moles of carbon dioxide produced

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.

Step 3: Determine the percentage by mass of magnesium carbonate in the mixture

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%.