6.4 g of CH4 is burned in 32 g of O2: CH4 + 2O2 -> CO2 + 2H2O. The percentage of O2 in excess over the stoichiometric requirement is (C = 12, H = 1, O = 16)

JEE Main levelSingle correct optionDifficulty 2 of 3mc191

$6.4 g$ of $CH X_{4}$ is burned in $32 g$ of $O X_{2}$ : $CH X_{4} + 2 O X_{2} CO X_{2} + 2 H X_{2} O$ . The percentage of $O X_{2}$ in excess over the stoichiometric requirement is (C = 12, H = 1, O = 16)

n(CH4) = 0.40 mol needs 0.80 mol O2; available n(O2) = 1.00 mol. Excess = 0.20 mol. Percent excess = 0.20/0.80 x 100 = 25%.