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 42- longrightarrow Mn 2++ CO 2+2 H 2 O

Question

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 42- longrightarrow Mn 2++ CO 2+2 H 2 O (A) 54.0 % (B) 27.1 % (C) 42 % (D) 84 %

Tardigrade Admin · Accepted Answer

M eq. of KMnO 4= M eq. of C 2 O 4-2= M eq. of CaCO 3 40 × .25= M eq. of CaO (10 × 10-3/2)= Mole of CaO % CaO =(5 × 10-3 × 56 × 100/.518)=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 $C a C O_{3}$

$40 \times .25 = M$ eq. of $C a O$

$\frac{10 \times 1 0 ^{- 3}}{2} =$ Mole of $C a O$

$% C a O = \frac{5 \times 1 0 ^{- 3} \times 56 \times 100}{.518} = 54%$

54.0%

27.1%

42%

84%

M eq. of KMnO4​=M eq. of C2​O4−2​=M eq. of CaCO3​ 40×.25=M eq. of CaO 210×10−3​= Mole of CaO %CaO=.5185×10−3×56×100​=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 $C a C O_{3}$

$40 \times .25 = M$ eq. of $C a O$

$\frac{10 \times 1 0 ^{- 3}}{2} =$ Mole of $C a O$

$% C a O = \frac{5 \times 1 0 ^{- 3} \times 56 \times 100}{.518} = 54%$