Determine the volume of gas that is released when 12.8 g of copper interacts with an excess of concentrated nitric acid

Let’s find the amount of copper substance by the formula:

n = m: M.

M (Cu) = 64 g / mol.

n = 12.8 g: 64 g / mol = 0.2 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

Cu + 4HNO3 (conc) = Cu (NO3) 2 + 2NO2 + 2H2O.

According to the reaction equation, 1 mol of copper accounts for 2 mol of nitrogen oxide. The substances are in quantitative ratios of 1: 2.

The amount of copper and nitrogen oxide will be equal.

n (NO2) = 2n (Cu) = 0.2 × 2 = 0.4 mol.

Let’s find the volume of the gas.

V = Vn n, where Vn is the molar volume of gas equal to 22.4 l / mol, and n is the amount of substance.

V = 0.4 mol × 22.4 L / mol = 8.96 L.

Answer: 8.96 liters.