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Chemistry

10 g. of a mixture of NaCl and anhydrous Na2SO4 is dissolved in water. An excess of BaCl2 soln. is added and 6.99 g. of BaSO4 is precipitated according to the equation :

Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl.

Calculate the percentage of Na2SO4 in the original mixture.

[O = 16 ; Na = 23 ; S = 32; Ba = 137]

Stoichiometry

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Answer

Na2SO4+BaCl2⟶BaSO4↓+2NaCl2(23)+32137+32+4(16)+4(16)=46+32=137+32+64+64=142 g=233 g\begin{matrix} \text{Na}2\text{SO}4 & + & \text{BaCl}2 & \longrightarrow & \text{BaSO}4 \downarrow & + & 2\text{NaCl} \ 2(23) + 32 & & & & 137 + 32 \ + 4(16) & & & & + 4(16) \ = 46 + 32 & & & & = 137 + 32 \ + 64 & & & & + 64 \ = 142 \text{ g} & & & & = 233 \text{ g} \ \end{matrix}

233 g of BaSO4 is obtained from 142 g of Na2SO4

6.99 g of BaSO4 will be obtained from 142233\dfrac{142}{233} x 6.99 = 4.26 g of Na2SO4.

∴ In 10 g mixture 4.2610\dfrac{4.26}{10} x 100 = 42.6% Na2SO4 is present.

Hence, 42.6% Na2SO4 is present in the original mixture

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