The reaction of 22.4 liters of hydrogen with 22.4 liters of oxygen gave 16 g of water. Calculate the percentage of practical yield.

2Н2 + О2 = 2Н2О The ratio of hydrogen to oxygen 2 to 1 n (H2) = v / 22.4 = 22.4 / 22.4 = 1 mol n (О2) = v / 22.4 = 22.4 / 22.4 = 1 mol We will calculate by hydrogen The ratio of hydrogen to water 2 to 2-> 1 to 1 Therefore, water was obtained 1 mol. m (H2O) = n * M (H2O) = 1 * 18 = 18 g of water could be obtained theoretically% = (mass obtained practically / theoretical value) * 100% = (16/18) * 100% = 88.8%