Question

When magnesium is placed in lead (II) sulphate solution, it reacts to form magnesium sulphate and lead. How much lead (II) sulphate would be required to react with 327 g of magnesium? (Molar mass of lead = 207 g/mol)

• A. 4128.4 g
• B. 1619.5 g
• C. 6868.4 g
• D. 2788. 4 g

Answer 1

The correct option is A 4128.4 g

$PbS{O}_4+Mg\to MgS{O}_4+Pb$
Given: mass of magnesium = 327 g
According to the reaction,
1 mole of $PbS{O}_4$ reacts with 1 mole of Mg.
Moles of Mg $=\frac{\text{given mass}}{\text{molar mass}}=\frac{327}{24}=13.625$ moles
13.625 moles of Mg will require 13.625 moles of $PbS{O}_4$
Mass of $PbS{O}_4$ required = $13.625\times$ Molar mass of $PbS{O}_4$
Mass of $PbS{O}_4$ required = $13.625\times 303$ g
Mass of $PbS{O}_4$ required = 4128.375 g = 4128.4 g