Determine the mass of the sediment formed by the interaction of 200 g of 40% sodium hydroxide solution with 50 g of iron susphate 2.
Find the mass of sodium hydroxide in the solution.
The mass fraction of a substance is calculated by the formula:
W = m (substance): m (solution) × 100%,
m (substance) = (m (solution) × W): 100%,
m (substance) = (200 g × 40%): 100% = 80 g.
Find the amount of sodium hydroxide substance by the formula:
M (NaOH) = 40 g / mol.
n = m: M.
n = 80 g: 40 g / mol = 2 mol.
M (FeSO4) = 150 g / mol.
n (FeSO4) = 50 g: 150 g / mol = 0.33 mol.
NaOH is in excess, we solve the problem due to the lack of FeSO4.
Let’s compose the reaction equation, find the quantitative ratios of substances.
2NaOH + FeSO4 = Fe (OH) 2 ↓ + Na2SO4.
According to the reaction equation, there is 1 mole of iron (II) hydroxide per 1 mole of iron sulfate.
Substances are in quantitative ratios 1: 1.
The amount of substance will be the same.
n (FeSO4) = n (Fe (OH) 2) = 0.33 mol.
Let’s find the mass of the sediment by the formula:
m = n × M,
M (Fe (OH) 2) = 90 g / mol.
m = 0.33 mol × 90 g / mol = 29.7 g.
Answer: 29.7 g.