A mixture of iron (III) and potassium sulfates weighing 400 g was dissolved in water and an excess

A mixture of iron (III) and potassium sulfates weighing 400 g was dissolved in water and an excess of sodium hydroxide was added. Dropped out 64.2 g of precipitate. Mass fraction of potassium sulfate in the initial mixture (%).

To solve this problem, we write down the given: m (mixture) = 400 g, as a result of interaction with sodium hydroxide (NaOH), a precipitate with a mass of 64.2 g was formed
Find: mass fraction of potassium sulfate (K2SO4) -?
Decision:
Let us write down the reaction equations
K2SO4 + 2NaOH = 2KOH + Na2SO4 – this reaction does not proceed, as a result of which no precipitate is formed.
Fe2 (SO4) 3 + NaOH = Fe (OH) 3 + Na2SO4 – Fe (OH) 3 precipitate is formed
Let’s arrange the coefficients.
Fe2 (SO4) 3 + 6NaOH = 2Fe (OH) 3 + 3Na2SO4
Let’s calculate the mass of iron (III) sulfate, which has entered into the reaction.
Above iron (III) hydroxide we write 64.2 g, and under iron (III) hydroxide we write its molar mass.
M (Fe (OH) 3) = 56 + (16 + 1) * 3 = 107 g / mol
Since 2 mol of iron (III) hydroxide was formed, you need 2 mol * 107 g / mol = 214 g
M (Fe2 (SO4) 3) = 56 * 2 + (32 + 16 * 4) * 3 = 400 g / mol
Above iron (III) sulfate, we write x g, and under iron (III) sulfate, its molar mass is 400 g / mol.
Let’s compose and solve the proportion
x = 400 * 64.2 / 214 = 120 g
Calculate the mass of potassium sulfate
m (K2SO4) = m (mixtures) – m (Fe2 (SO4) 3)
m (K2SO4) = 400 – 120 = 280 g
Let’s write down the formula for calculating the mass fraction:
w = m (substances) / m (mixtures)
w (K2SO4) = 280/400 = 0.7 or 70%
Answer: w (K2SO4) = 0.7 or 70%