Black CuS is warmed with dilute HNO3 according to 3CuS + 8H+ + 2NO3- -> 3Cu^2+ + 3S + 2NO + 4H2O. The amount of NO liberated from 0.30 mol of CuS is

JEE Main levelSingle correct optionDifficulty 2 of 3qc030

Black $CuS$ is warmed with dilute $HNO X_{3}$ according to $3 CuS + 8 H X^{+} + 2 NO X_{3} X^{-} 3 Cu X^{2 +} + 3 S + 2 NO + 4 H X_{2} O$ . The amount of $NO$ liberated from $0.30 mol$ of $CuS$ is

The equation gives

2 mol

NO

for every

3 mol

CuS

. Hence

n (NO) = 0.30 \times 2/3 = 0.20 mol

. Sulphide is oxidised to elemental sulphur and nitrogen is reduced from

+ 5

+ 2