How many grams of sediment is formed when 500 kg of iron (II) sulfate, containing 20% impurities

| July 7, 2021 | education

How many grams of sediment is formed when 500 kg of iron (II) sulfate, containing 20% impurities, reacts with sodium hydroxide?

Let’s write the reaction equation:
FeSO4 + 2NaOH = Fe (OH) 2 + Na2SO4
First, let’s determine the mass of pure iron (II) sulfate:
We have 80% pure substance, 20% impurity.
Let’s write an expression to determine the mass fraction of a substance:
w (in-va) = (m (in-va) / m (mixture)) * 100%
m (in-va) = w (in-va) * m (mixture) / 100
Substitute the numerical values:
m (in-va) = w (in-va) * m (mixture) / 100 = 80 * 500/100 = 400 kg.
Based on the condition that the amount of substance according to the equation is the same:
m (FeSO4) / M (FeSO4) = m (Fe (OH) 2) / M (Fe (OH) 2)
Let’s define the molar masses:
M (FeSO4) = 56 + 32 + 16 * 4 = 152 g / mol
M (Fe (OH) 2) = 56 + 2 * (16 + 1) = 90 g / mol
Let us express m (Fe (OH) 2) from the formula:
m (Fe (OH) 2) = m (FeSO4) * M (Fe (OH) 2) / M (FeSO4)
Substitute the numerical values:
m (Fe (OH) 2) = m (FeSO4) * M (Fe (OH) 2) / M (FeSO4) = 400 * 90/152 = 236.8 kg.
Answer: the mass of the sediment is 236.8 kg.



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