Barium chloride was added to 75 g of 50% magnesium sulfate, how many g of the precipitate was formed.

1. We write down the reaction equation, since the sulfate ion is a high-quality reagent for barium ions, the result is a precipitate of BaSO4 ↓:

MgSO4 + BaCl2 = BaSO4 ↓ + MgCl2;

2.m (MgSO4 substance) = m (MgSO4 solution) * w (in fractions) = 75 * 0.5 = 37.5 g;

3.n (MgSO4) = m (MgSO4): M (MgSO4);

4. M (MgSO4) = 24 + 32 + 4 * 16 = 120 g / mol;

5.n (MgSO4) = 37.5: 120 = 0.3125 mol;

6.the amounts of magnesium and barium sulfates are in a ratio of 1: 1, therefore:

n (BaSO4) = 0.3125 mol;

7.M (BaSO4) = 137 + 32 + 4 * 16 = 233 g / mol

8.m (BaSO4) = n * M = 0.3125 * 233 = 72.81 g.

Answer: 72.81 g.