4 mol of H2 is mixed with 3 mol of O2 for 2H2 + O2 -> 2H2O. The percentage of O2 in excess over the amount required is

NEET levelSingle correct optionDifficulty 1 of 3mc187

$4 mol$ of $H X_{2}$ is mixed with $3 mol$ of $O X_{2}$ for $2 H X_{2} + O X_{2} 2 H X_{2} O$ . The percentage of $O X_{2}$ in excess over the amount required is

O2 required = 4/2 = 2 mol. Excess O2 = 3 - 2 = 1 mol. Percent excess = (1/2) x 100 = 50% (relative to the amount required).

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