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Question
When excess lead nitrate solution was added to a solution of sodium sulphate, 15.1g of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
Na2SO4 + Pb(NO3)2 → PbSO4 + 2NaNO3
(H = 1, C = 12, O = 16, Na = 23, S = 32, Pb = 207)
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Solution
Na2SO4 + Pb(NO3)2 → PbSO4 + 2NaNO3
molecular weight of Na2SO4 is 142g
molecular weight of PbSO4 is 303g
303g of PbSO4 is formed by 142g of Na2SO4
15.1g of PbSO4 is formed by = 142 X 15.1/303 = 7.1g of Na2SO4.
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