For 4NH3 + 5O2 -> 4NO + 6H2O, 3.4 g of NH3 is mixed with 11.2 g of O2. Find the percentage of O2 in excess over the amount required for complete reaction of NH3, to the nearest integer (N = 14, H = 1, O = 16).

JEE Main levelNumerical answerDifficulty 2 of 3mc194

For $4 NH X_{3} + 5 O X_{2} 4 NO + 6 H X_{2} O$ , $3.4 g$ of $NH X_{3}$ is mixed with $11.2 g$ of $O X_{2}$ . Find the percentage of $O X_{2}$ in excess over the amount required for complete reaction of $NH X_{3}$ , to the nearest integer (N = 14, H = 1, O = 16).

Numerical answer type, answer in %, tolerance 0.5.

n(NH3) = 0.20 mol needs 0.200 x 5/4 = 0.25 mol O2. Available O2 = 11.2/32 = 0.35 mol. Excess = 0.10 mol, which is 0.10/0.25 x 100 = 40%.