# When processing technical copper, containing 4% of inert impurities, diluted with nitric acid

**When processing technical copper, containing 4% of inert impurities, diluted with nitric acid, 33.6 liters of gas were released. Determine the mass of technical copper**

Given:

m (NO) = 18 g.

V (gas) -?

Solution:

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

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

n = V: Vn

V = 33.6 L: 22.4 L / mol = 1.5 mol.

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

3 Cu + 8HNO3 (diluted) = 3Cu (NO3) 2 + 2NO2 ↑ + 4H2O

According to the reaction equation, there is 3 mol of copper for 2 mol of gas. Substances are in quantitative ratios of 2: 3.

The amount of copper substance will be 1.5 times more than the amount of nitric oxide substance.

n (Cu) = 1.5 mol × 1.5 = 2.25 mol.

Or :

3 mol (Cu) – 2 mol (NO2)

x mol (Cu) – 1.5 mol (NO2).

x mol (Cu) = (1.5 mol × 3 mol): 2 mol = 2.25 mol.

Let’s find the mass of copper.

m = n M

M (Cu) = 64 g / mol.

m = 2.25 mol × 64 g / mol = 144 g.

Let’s find the mass of copper without impurities.

100% – 4% = 96%.

144 g – 100%,

m – 96%,

m = (144 g × 96%): 100% = 138.24 g.

Answer: 138.24 g.