The interaction of calcium chloride weighing 40 g with sodium sulfate precipitated a precipitate weighing 4 g.

| March 20, 2021 | education

The interaction of calcium chloride weighing 40 g with sodium sulfate precipitated a precipitate weighing 4 g. Determine the yield of the reaction product.

1. Let’s compose the equation of the proceeding reaction:

CaCl2 + Na2SO4 = CaSO4 ↓ + 2NaCl;

2.Calculate the chemical amount of calcium chloride:

n (CaCl2) = m (CaCl2): M (CaCl2);

M (CaCl2) = 40 + 2 * 35.5 = 111 g / mol;

n (CaCl2) = 40: 111 = 0.36 mol;

3. Determine the theoretical amount of calcium sulfate:

ntheor (CaSO4) = n (CaCl2) = 0.36 mol;

4. Let us establish the practical chemical amount of the sediment:

npr (CaSO4) = m (CaSO4): M (CaSO4);

M (CaSO4) = 40 + 32 + 4 * 16 = 136 g / mol;

npr (CaSO4) = 4: 136 = 0.03 mol;

5. calculate the yield of the reaction product:

ν = npr (CaSO4): ntheor (CaSO4) = 0.03: 0.36 = 0.0833 or 8.33%.

Answer: 8.33%.



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