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Stoichiometric Calculations

0.518 g sampl...

Question

$0.518$ g sample of limestone is dissolved and calcium is precipitated as calcium oxalate $(C a C_{2} O_{4})$ . After filtration and washing the precipitate, it requires $40$ mL of $0.05 M K M n O_{4}$ solution acidified with sulphuric acid to titrate it. What is the percentage of $C a O$ in the limestone? (as nearest integer)

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Solution

$40$ mL of $0.05 M {M n O_{4}}^{-}$ corresponds to $40$ mL of $0.05 \times 5 N {M n O_{4}}^{-}$ .
This is equal to $10$ milliequivalents or $0.01$ equivalents.
$0.01$ equivalents of oxalate ion is present.
$0.01$ equivalents of $C a O$ corresponds to $0.005$ moles or $0.005 56$ g $= 0.28$ g
This is present in $0.518$ g sample of limestone.
The percentage of $C a O$ in the limestone sample is $\frac{0.28}{0.518} \times 100 = 54 %$ .

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