Question

Na2SO4 + Pb(NO3)2 ⟶ PbSO4 + 2NaNO3. When excess lead nitrate solution was added to a solution of sodium sulphate, 15.15 g of lead sulphate were precipitated. What mass of sodium sulphate was present in the original solution?
[H = 1 ; C = 12 ; O = 16 ; Na = 23; S = 32 ; Pb = 207]

August 5, 2024

Answer

Na₂SO₄ + Pb(NO₃)₂ → PbSO₄+ 2NaNO₃

142 g 303 g

303 g of lead sulphate was obtained from 142 g of Na₂SO₄

∴ 15.15 g of lead sulphate will be precipitated from $142\over303$ × 15.15 = 7.1 g

Hence, 7.1 g of sodium sulphate was present in the original solution.

💡 Some Related Questions

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