Question

What mass of copper is required to produce 13.0 L of NO at 725 torr and 20.0 ° C? Nitric oxide is produced in the reaction between copper metal and nitric acid: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 4 H2O(l) + 2 NO(g)

Answer 1

Calculate the number of moles of NO generated under the said conditions.

P = 725 torr = (725 torr)*(1 atm/760 torr) = 0.9539 atm (1 atm = 760 torr); V = 13.0 L and T = 20.0°C = (20.0 + 273) K = 293.0 K.

Use the ideal gas law to find n, the number of moles.

P*V = n*R*T where R = 0.082 L-atm/mol.K.

Therefore,

(0.9539 atm)*(13.0 L) = n*(0.082 L-atm/mol.K)*(293.0 K)

===> n = (0.9539*13.0)/(0.082*293.0) mole = 0.5161 mole.

Moles of NO generated = 0.5161 mole.

As per the balanced stoichiometric equation,

2 moles NO = 3 moles Cu.

Therefore,

0.5161 mole NO = (0.5161 mole NO)*(3 moles Cu/2 moles NO) = 0.77415 mole.

Molar mass of Cu = 63.546 g/mol.

Therefore, mass of Cu required = (63.546 g/mol)*(0.77415 mole) = 49.1941 g/mol ≈ 49.19 g(ans).