A 0.518 g sample of limestone is dissolved in HCl and then the calcium is precipitated as CaC2O4. After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO4 solution acidified with H2SO4 to titrate it as. The percentage of CaO in the sample isMnO4- + H+ + C2O4-2 → Mn2+ +CO2 + 2H2O

A 0.518 g sample of limestone is dissolved in HCl and then the calcium is precipitated as ${CaC}_{2} O_{4}$ . After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N ${KMnO}_{4}$ solution acidified with $H_{2} {SO}_{4}$ to titrate it as. The percentage of CaO in the sample is
${MnO}_{4}^{-} + H^{+} + C_{2} O_{4}^{- 2} \overset{}{\to} {Mn}^{2 +} + {CO}_{2} + 2 H_{2} O$

A
54.0%
B
27.1%
C
42%
D
84%

Solution:

M eq. of ${KMnO}_{4}$ = M eq. of $C_{2} O_{4}^{- 2}$ = M eq. of ${CaCO}_{3}$

40 x .25 = Meq of CaO

$\frac{10 \times 10^{- 3}}{2} = Mole of CaO % CaO = \frac{5 \times 10^{- 3} \times 56 \times 100}{. 518} CaO = 54 %$

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