Find the mass of the precipitate obtained by the reaction of 40 g of potassium hydroxide and 60 g of ferrous sulfate (2).

2KOH + FeSO4 = K2SO4 + Fe (OH) 2.
Algorithm for solving the problem:
1) Determination of the molar mass of potassium hydroxide.
2) Determination of the molar mass of ferrous sulfate.
3) Calculation of the amount of potassium hydroxide substance.
4) Calculation of the amount of ferrous sulfate substance.
5) Definition of excess – lack.
6) Calculation of the amount of sediment substance.
7) Calculation of the mass of the sediment.
Solution:
1) 39 + 16 + 1 = 56.
2) 56 + 32 + 4 * 16 = 152.
3) 40/56 = 0.7 mol.
4) 60/152 = 0.4 mol.
5) Iron sulfate is in short supply. The calculation is carried out on it.
6) 0.4 mol.
7) 0.4 * (56 + 32 + 2) = 36g – the answer.