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Chemistry

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]

Stoichiometry

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Answer

Na2SO4+Pb(NO3)2⟶PbSO4+2NaNO32(23)+32207+32+4(16)+4(16)=46+32=207+32+64+64=142 g=303 g\begin{matrix} \text{Na}2\text{SO}4 & + & \text{Pb}(\text{NO}3)2 & \longrightarrow &\text{PbSO}4 & + & 2\text{NaNO}3 \ 2(23) + 32 & & & & 207 + 32 \ + 4(16) & & & & + 4(16) \ = 46 + 32 & & & & = 207 + 32 \ + 64 & & & & + 64 \ = 142 \text{ g} & & & & = 303 \text{ g} \ \end{matrix}

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

∴ 15.15 g of lead sulphate will be precipitated from 142303\dfrac{142}{303} x 15.15 = 7.1 g.

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

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