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
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B
1619.5 g
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C
6868.4 g
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D
2788. 4 g
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Solution
The correct option is A 4128.4 g PbSO4+Mg→MgSO4+Pb
Given: mass of magnesium = 327 g
According to the reaction,
1 mole of PbSO4 reacts with 1 mole of Mg.
Moles of Mg =given massmolar mass=32724=13.625 moles
13.625 moles of Mg will require 13.625 moles of PbSO4
Mass of PbSO4 required = 13.625× Molar mass of PbSO4
Mass of PbSO4 required = 13.625×303 g
Mass of PbSO4 required = 4128.375 g = 4128.4 g