Question

5 g of $K_{2}S{O}_4$ were dissolved in 250 mL of solution. How many mL of this solution should be used so that 1.2 g of $BaS{O}_4$ may be precipitated from $BaC{l}_2$ solution ?

Answer 1

The desired equation is:

$BaC{l}_2+\underset{=174g}{\underset{2\times 39+32+64}{K_{2}S{O}_4}}\to \underset{=233g}{\underset{137+32+64}{BaS{O}_4}}+2KCl$

233 g of $BaS{O}_4$ obtained from 174 g of $K_{2}S{O}_4$

1.2 g of $BaS{O}_4$ will be obtained from $\frac{174}{233}\times 1.2$

$=0.8961gof{K}_{2}S{O}_4$

5g of $K_{2}S{O}_4$ are present in 250 mL of solution

So, 0.8961 g of $K_{2}S{O}_4$ will be present in $\frac{250}{5}\times 0.8961$

$=44.8mL$ of solution