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 is: MnO-4 + H+ + C2O42 → Mn2+ + CO2 + 2H2O

Question

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 is: MnO-4 + H+ + C2O42 → Mn2+ + CO2 + 2H2O (A) 54.0 % (B) 27.1 % (C) 42 % (D) 84 %

Tardigrade Admin · Accepted Answer

M eq. of KMnO4 = M eq. of C2O42- = M eq. of CaCO3 40 × .25 = Meq of CaO (10 ×10-3/2) = mol of Cao % Cao =(5×10-3 ×56×100/.518) Cao =54 %

$54.0%$

$27.1%$

$42%$

$84%$

M eq. of $K M n O_{4} =$ M eq. of $C_{2} O_{4}^{2 -} =$ M eq. of $C a C O_{3}$
$40 \times .25 =$ Meq of $C a O$
$\frac{10 \times 1 0 ^{- 3}}{2} =$ mol of $C a o$
$% C a o = \frac{5 \times 1 0 ^{- 3} \times 56 \times 100}{.518}$
$C a o = 54%$

54.0%

27.1%

42%

84%

M eq. of KMnO4​= M eq. of C2​O42−​= M eq. of CaCO3​ 40×.25= Meq of CaO 210×10−3​= mol of Cao %Cao=.5185×10−3×56×100​ Cao=54%

$54.0%$

$27.1%$

$42%$

$84%$

M eq. of $K M n O_{4} =$ M eq. of $C_{2} O_{4}^{2 -} =$ M eq. of $C a C O_{3}$
$40 \times .25 =$ Meq of $C a O$
$\frac{10 \times 1 0 ^{- 3}}{2} =$ mol of $C a o$
$% C a o = \frac{5 \times 1 0 ^{- 3} \times 56 \times 100}{.518}$
$C a o = 54%$