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Stoichiometric Calculations

5 g of K2SO...

Question

$5$ g of $K_{2} S O_{4}$ was dissolved in $250$ mL of solution. The amount of mL of this solution that should be used so that $1.2$ g $B a S O_{4}$ may be precipitated by this solution on addition of $B a {C l}_{2}$ solution is __________. (as the nearest integer)

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Solution

Meq. of $B a S O_{4}$ formed $= \frac{1.2}{\frac{233}{2}} \times 1000 = 10.30$
$∴$ Meq. of $K_{2} S O_{4}$ required $= 10.30$
Also, normality of given $K_{2} S O_{4} =$ $\frac{5 \times 1000}{\frac{174}{2} \times 250}$
Volume of $K_{2} S O_{4}$ solution can be obtained as follows:
$M e q . = N \times V$
$⟹ 10.30 = \frac{5 \times 1000}{\frac{174}{2} \times 250} \times V$
Therefore, $V = 44.8$ L
So, the nearest integer is $45$ .

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K_{2} S O_{4}

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