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.

Short answer

The balanced chemical equations for the reactions between hydrochloric acid and magnesium carbonate, and between hydrochloric acid and calcium carbonate, are: $${\mathrm{MgCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})⟶{\mathrm{MgCl}}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{CO}}_2(\mathrm{g})$$ $${\mathrm{CaCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})⟶{\mathrm{CaCl}}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{CO}}_2(\mathrm{g})$$ Approximately 0.0725 moles of CO2 are produced from the reactions. The percentage by mass of magnesium carbonate in the mixture is approximately 46.8%.

Step-by-step solution

Write balanced chemical equations for the reactions

For the reaction between hydrochloric acid (HCl) and magnesium carbonate (MgCO3), the balanced chemical equation is given by: $${\mathrm{MgCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})⟶{\mathrm{MgCl}}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{CO}}_2(\mathrm{g})$$ For the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3), the balanced chemical equation is given by: $${\mathrm{CaCO}}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})⟶{\mathrm{CaCl}}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{CO}}_2(\mathrm{g})$$

Calculate the total number of moles of carbon dioxide formed

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 ($0.0821\frac{\mathrm{atm}·\mathrm{L}}{\mathrm{K}·\mathrm{mol}}$) T = temperature in K Rearrange the formula to find n: $$n=\frac{PV}{RT}$$ Plugging in the values, we get: $$n=\frac{(0.9776\mathrm{atm})(1.72\mathrm{L})}{(0.0821\frac{\mathrm{atm}·\mathrm{L}}{\mathrm{K}·\mathrm{mol}})(301.15\mathrm{K})}$$ $$n\approx 0.0725\mathrm{mol}$$ There are approximately 0.0725 moles of CO2 produced from the reaction.

Calculate the percentage by mass of magnesium carbonate in the mixture

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 = $\frac{3.056\mathrm{g}}{6.53\mathrm{g}}\times 100\approx 46.8\mathrm{\%}$ The percentage by mass of magnesium carbonate in the mixture is approximately 46.8%.