chemistry question

Question

Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water.

N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)

If 3.05 g of N2H4 reacts with excess oxygen and produces 0.450 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

Find theoretical yield:

(3.05 g N2H4)(1 mol N2H4/32.045 g N2H4)(1 mol N2/ 1 mol N2H4) ( 28.01 g N2/ 1 mol N2) = 2.67 grams N2

Actual yield:

use PV = nRT, solve for n

--> (1.00 atm)(0.450L) = n(0.0821 L atm / mol K)(295 K)

--> n = 0.01858 mol N2

--> convert to grams: 0.01858 mol N2 ( 28.01 g N2 / 1 mol) = 0.5204 g N2

Percent yield:

actual / theoretical --> 0.5204 g / 2.67 g = 19.5% yield