Moles can be produced/which reactant is limiting/percent yield of the reaction

Question

Ammonia, NH3, may react with oxygen to form nitrogen gas and water.

If 3.55 g of NH3 reacts with 5.33 g O2 and produces 0.550 L of N2, at 295 K and 1.00 atm, how many moles of N2 can be produced by each reactant?

Based on the theoretical yields above, which reactant is limiting?

What is the percent yield of the reaction?

Answer 1

Hello Ma

The balanced chemical reaction is:

4NH3 (g) + 3O2 (g) --> 2N2 (g) + 6H2O (l)

1) First convert g of NH3 and O2 to moles of N2. The reactant that produces the lowest # of moles of N2 is the limiting reagent.

3.55 g NH3 x (1 mol NH3 / 17 g NH3) x (2 mol N2 / 4 mol NH3) = 0.104 mol N2

5.33 g O2 x (1 mol O2 / 32 g O2) x (2 mol N2 / 3 mol O2) = 0.111 mol N2

NH3 produces less N2, therefore, it is the limiting reagent.

2) Use the ideal gas law equation to calculate the # of moles of N2.

PV = nRT

n = PV / RT

P = 1.00 atm

V = 0.550 L

R = 0.0821 L atm / mol K

T = 295 K

n = (1.00 x 0.550) / (0.0821 x 295) = 0.0227 mol N2

%-Yield = (0.0227 mol / 0.104 mol) x 100% = 21.8%