Hydrogen sulfur with a mass of 120 g, containing 2% of impurities, reacts with copper sulfate2

Hydrogen sulfur with a mass of 120 g, containing 2% of impurities, reacts with copper sulfate2 with a mass of 200 g. find the mass of the resulting residue.

m (H2S) = 120 g.

W (impurities) = 2%.

m (CuSO4) = 200 g.

Determine the mass of the resulting sediment. We write down the solution.

H2S + CuSO4 = CuS + H2SO4

First, we find a mass of pure hydrogen sulfide.

100% – 2% = 98%.

120 g – 100%.

X g – 98%.

X = 98 * 120: 100 = 117.6 g.

We find the mass of barium sulfate, which should be.

M (H2S) = 1 * 2 + 32 = 34 g / mol.

M (CuSO4) = 64 + 32 + 16 * 4 = 160 g / mol.

117.6 g H2S – x g CuSo4

34 g / mol H2S – 160 g / mol CuSO4

X = 553 g.

This means that there should be 553 g of CuSo4. Consequently, copper sulfate is in short supply. Calculated by copper sulfate.

M (CuS) = 64 + 32 = 96 g / mol.

200 g CuSo4 – x g CuS

160 g / mol CuSO4 – 96 g / mol CuS

X = 200 * 96: 160 = 120 g CuS.

Answer: 120 g.