Answer to Question #52333 in Inorganic Chemistry for Cody Filingeri
Calculate the % yield from an experiment that produced 5.85 grams of Nitrogen gas from the combustion of 66.92 grams of CH3NH2 with 79.8 grams of O2 at 273°K and 1 ATM CH3NH2(g) + O2(g) → CO2 + N2 + H2O.
The balanced equation: 4CH3NH2(g) + 9O2(g) → 4CO2 + 2N2 + 10H2O First we need to determine the limiting reactant: The moles of CH3NH2 = 66.92/31g/mole = 2.16 moles The moles of O2 = 79.8/32g/mole = 2.49 moles So, the limiting reactant is CH3NH2. The ratio of CH3NH2/N2 is 2. So, the number of moles of N2 produced will be 2.16/2 = 1.08 moles. The mass of N2 = 1.08*28g/mole = 30.24g. The % yield is (5.85/30.24)*100% = 19.35%