When 9.0 g of hydrogen gas, H2, reacts with oxygen gas, O2, 73.0 g of water is produced.
a.) What is the theoretical yield of water?
b.) What is the percentage yield?
Theequation for this reaction is 2H2 + O2 = 2H2O; Having a mass of hydrogen we can find the quantity substance n(H2) = m(H2)/M(H2); M(H2) = 1 *2 = 2 g/mol n(H2) = 9 g / 2 g/mol = 4.5 moles; The amount of substance of hydrogen with the amount of substance of water as a 2 to 2 or 1 to 1. so n(H2O) = 4.5 moles Theoretical mass of water will be n(H2O) = m(H2O)/M(H2O); M(H2O) = 1 *2 + 16 = 18 g/mol m (H2O) 4.5 moles * 18 g/mol = 81 g. so the theoretical yield of water is 81 g. The percentage yield will be m(H2O)practical / m(H2O)theoretical = 73.0 g / 81.0 g = 0.9012 or 90.12 %