When 10 g of technical zinc was dissolved in an excess of dilute hydrochloric acid, 3.1

When 10 g of technical zinc was dissolved in an excess of dilute hydrochloric acid, 3.1 l of hydrogen was released, determine the mass fraction of impurities in this sample of zinc.

1. Let’s compose the equation for the reaction of zinc with hydrochloric acid:

Zn + 2HCl = ZnCl2 + H2 ↑;

2. Let’s calculate the chemical amount of released hydrogen:

n (H2) = V (H2): Vm = 3.1: 22.4 = 0.1384 mol;

3. Set the amount of dissolved zinc:

n (Zn) = n (H2) = 0.1384 mol;

4. Find the mass of zinc:

m (Zn) = n (Zn) * M (Zn) = 0.1384 * 65 = 9 g;

5. Determine the mass of impurities:

m (impurities) = m (sample) – m (Zn) = 10 – 9 = 1 g;

6. Let’s calculate the mass fraction of impurities:

w (impurities) = m (impurities): m (sample) = 1: 10 = 0.1 or 10%.

Answer: 10%.