To a solution containing 39 g of sodium sulfide was added a solution containing 80 g

| February 5, 2021 | education

To a solution containing 39 g of sodium sulfide was added a solution containing 80 g of zinc chloride. Calculate the mass of the precipitate formed.

Find the amount of sodium sulfide Na2S by the formula:

n = m: M.

M (Na2S) = 78 g / mol.

n = 39 g: 78 g / mol = 0.5 mol.

M (ZnCl2) = 136 g / mol.

n = 80: 136 = 0.588 mol.

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

Na2S + ZnCl2 = ZnS ↓ + 2NaCl.

Na2S is given in deficiency.

According to the reaction equation, there is 1 mole of ZnS per 1 mole of Na2S.

The substances are in quantitative ratios of 1: 1.

The amount of substance will be equal.

n (ZnS) = n (ZnCl2) = 1: 1 = 0.5 mol.

Find the mass of ZnCl2

M (ZnS) = 97 g / mol.

m = n × M.

m = 97 g / mol × 0.5 mol = 48.5 g.

Answer: 48.5 g.



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