Determine the mass fractions (in percent) of iron and zinc sulfides in the mixture if, when processing 28.2 g

Determine the mass fractions (in percent) of iron and zinc sulfides in the mixture if, when processing 28.2 g of this mixture with an excess of hydrochloric acid solution, a gas is released, which completely precipitates copper from 405 g of a 10% solution of copper (II) chloride.

1. Let’s write down the equations of the proceeding reactions:

FeS + 2HCl = FeCl2 + H2S ↑;

ZnS + 2HCl = ZnCl2 + H2S ↑;

H2S + CuCl2 = CuS ↓ + 2HCl;

2.Calculate the mass of copper chloride (2):

m (CuCl2) = w (CuCl2) * m (solution);

m (CuCl2) = 0.1 * 405 = 40.5 g;

3. find the chemical amounts of copper chloride and hydrogen sulfide:

n (CuCl2) = m (CuCl2): M (CuCl2);

M (CuCl2) = 64 + 35.5 * 2 = 135 g / mol;

n (CuCl2) = 40.5: 135 = 0.3 mol;

n (H2S) = n (CuCl2) = 0.3 mol;

4.Let the chemical quantities:

n (FeS) = x, and n (ZnS) = y, since according to the reaction n (FeS) = n (H2S) and n (ZnS) = n (H2S), then:

x + y = 0.3 mol;

5.we express the masses of the initial sulfides and substitute them into the mass of the mixture:

m (FeS) = n (FeS) * M (FeS) = 88 * x;

m (ZnS) = n (ZnS) * M (ZnS) = 97 * y;

88 * x + 97 * y = 28.2 g.

6. make up a system of equations and find the chemical amounts of iron and zinc sulfides:

x + y = 0.3;

88 * x + 97 * y = 28.2;

y = 0.3 – x;

88 * x + 97 * (0.3 – x) = 28.2;

9 * x = 0.9;

x = 0.1 mol;

y = 0.2 mol;

7.Calculate the masses of sulfides:

m (FeS) = 0.1 * 88 = 8.8 g;

m (ZnS) = 0.2 * 97 = 19.4 g;

8.determine the mass fractions:

w (FeS) = m (FeS): m (mixtures) = 8.8: 28.2 = 0.3121 or 31.21%;

w (ZnS) = m (ZnS): m (mixtures) = 19.4: 28.2 = 0.6879 or 68.79%.

Answer: 31.21% FeS; 68.79% ZnS.