Use Cu = 63.5 and molar volume 22.4 L/mol at STP. 6.35 g of Cu reacts completely with nitric acid by 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O (dilute) or Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O (concentrated). Select the correct statement(s).

Dilute acid gives 1.49 L of NO at STP

Concentrated acid gives 4.48 L of NO2 at STP

Concentrated acid consumes 0.40 mol of HNO3

Dilute acid gives 1.49 L of NO at STP

Concentrated acid gives 4.48 L of NO2 at STP

Concentrated acid consumes 0.40 mol of HNO3

Dilute acid consumes 0.20 mol of HNO3

JEE Main levelOne or more correct optionsDifficulty 2 of 3rx143

Use Cu = 63.5 and molar volume $22.4 L/mol$ at STP. $6.35 g$ of Cu reacts completely with nitric acid by $3 Cu + 8 HNO X_{3} 3 Cu (NO X_{3}) X_{2} + 2 NO + 4 H X_{2} O$ (dilute) or $Cu + 4 HNO X_{3} Cu (NO X_{3}) X_{2} + 2 NO X_{2} + 2 H X_{2} O$ (concentrated). Select the correct statement(s).

n (Cu) = 0.1

. Dilute:

NO = 0.1 \times 2/3

mol, i.e. 1.49 L (A true);

HNO X_{3} = 0.1 \times 8/3 = 0.267

mol, so D (0.20) is false. Concentrated:

NO X_{2} = 0.2

mol, i.e. 4.48 L (B true);

HNO X_{3} = 0.1 \times 4 = 0.40

mol (C true).