Question

4. Carbon dioxide can be generated by the reaction of hydrochloric acid with calcium carbonate. How many milliliters of dry CO, at 20.0 °C and 755 torr can be prepared from a mixture of 32.1 g of Caco, and 203 mL of 0.245 M HCI?

Answer 1

Write the reaction as follows: CaCO3 + 2HC1 —> CaC1, +CO, +H2O 1 mole of Caco reacts with 2 moles of HC1 gives 1 mole of CaCl2, 1 mole of Co, and 1 mole of H,O. Mass of CaCO3 = 32.18 molar mass of CaCO3 = 100.0 g/mol Mass of Caco moles of Caco, = molar mass of CaCO3 32.1g = 32. 1x10-mol 100.0 g/mol Molarity of HC1 = 0.245 M volume of HC1 = 203 mL moles of HC1 = Molarity of HC1 x volume of HC1 = 0.245 Mxmol/Lx203 mx _IL 1000 ml = 49.7x10-mol M From the reaction the stoichiometric relation between Caco, and HCl is moles of Caco moles of HCI Substitute the calculated moles in the above relation, the moles with least is the limitig reagent and the moles with highest is the excess reagent. 32.1x10-mol Caco 49.7x10-mol HCI 1 --- 32.1x10-2 mol Caco, > 24.85x10-3 mol HCI So, the limiting reagent is HC1. Therefore, the limiting reagent is Ha --------------------------------- ----------- Calculate the moles of CO, from the limiting reactant as follows: From the reaction the stoichiometric relation between CO, and HCl is moles of Co_moles of HC1 2 49.7x10 mol moles of CO2, n=9 - 1 »= 24.85x10?mol — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — Vapor pressure of water is 17.5 torr and the total pressure is 755 torr. total pressure = Parti al pressure of CO2 + vapor pressure of water Parti al pressure of dry CO, = total pressure - vapor pressure of water Pco, = 755 torr –17.5 torr = 737.5 torr Ideal gas equation, PV = nRT ....... (1) gas cosntant, R= 62.363 L torr/molK Temperature, T = 20.0°C= 20.0+273 K = 293K Substitute the correspondig valuesin equation (1) and calculate the volume, V. V = nRT TE 24.85x10-mol X62.363 L torr/molKx293K 737.5 torr V = 0.616 Lx 1000 mL V = 616 mL Therefore, the requried solution is 616 mL)