ANSWERS: 1

225 g of CaCO3, with a molar mass of 100.0869 g/mol is 2.24805 moles. Since the equation is balanced and the coefficients of all reactants and products is one, the molar ratio of CO2:CaCO3 in the reaction should be 1:1. There ought to also be 2.24805 moles of CO2. CO2 has a molar volume of 22.4 L/mol at STP, so, you should, theoretically, be able to produce 50.356 L of CO2 in the reaction. Since you only recovered 28.2 L, you only achieved a 56% yield.
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