Stoichiometry: yield of reaction between hydrochloric acid and calcium carbonate

 P: 205 Thank you in advance for the correction of the resolution. 1. The problem statement, all variables and given/known data By reacting 225 g of hydrochloric acid with 300 g of calcium carbonate with 80% of purity, 34 L (litres) of gas were obtained, measured at 37oC and a pression of 0.82 atm. What is the yield of this reaction and what mass of calcium carbonate reacted? (Remember that H2CO3, in water, decomposes as water and CO2) $$2HCl_{(aq)} + CaCO_3_{(s)} \rightarrow CaCl_2_{(s)} + H_2CO_3_{(aq)}$$ R = 0.082 (atm*L)/(mol*K). 2. Relevant equations $$PV = nRT$$ 3. The attempt at a solution $$2HCl_{(aq)} + CaCO_3_{(s)} \rightarrow CaCl_2_{(s)} + H_2O_{(aq)} + CO_2_{(g)}$$ Molecular masses: 2 HCl = 2(36.5) = 73 g; 1CaCO3 = 100 g. Assuming that the expression "with 80% of purity" refers to calcium carbonate (we are not sure): The molar relation in order to know the limiting reagent (but the purity of the calcium carbonate is 80%, so 300 g * 80% = 240 g): $$1 mol HCl \rightarrow 1 mol CaCO_3$$ $$73 g \rightarrow 100g$$ $$x \rightarrow 300 \times 0.8$$ x = 175.2 g of HCl. Smaller than the available quantity, 225 g, thus it is in excess. The limiting reagent is CaCO3. Now, the molar relation between calcium carbonate and the gas, CO2, in order to find how many litres of the gas (y) are produced with 100% yield. But first, to find the molar volume: 37oC = 37 + 273 = 310 K. $$PV_m = nRT \rightarrow 0.82V_m = 0.082 \times 310 \rightarrow V_m = 31 L$$ $$1 mol CaCO_3 \rightarrow 1 mol CO_2$$ $$100 g \rightarrow 31L$$ $$240 g \rightarrow y$$ y = (31 * 240) / 100 y = 74.4 L. What is the yield of this reaction? Since the volume of gas produced with the current yield is 34 L, the yield (R) is: $$R = \frac{34}{74.4}$$ R = 45.6%. What mass of calcium carbonate reacted? It was already calculated: 240 g.