Answer to Question #52333 in Inorganic Chemistry for Cody Filingeri

Question

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.

Answer 1

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%