Moles of KHC 2 O 4 (potassium acid oxalate) required to reduce 100 mL of 0.02 M KMnO 4 in acidic medium (. to .Mn 2+) is

Question

Moles of KHC 2 O 4 (potassium acid oxalate) required to reduce 100 mL of 0.02 M KMnO 4 in acidic medium (. to .Mn 2+) is (A) 0.002 (B) 0.005 (C) 0.001 (D) 0.007

Tardigrade Admin · Accepted Answer

5 C 2 O 42-+2 MnO 4- arrow Mn 2++10 CO 2 2 mol of MnO 4- ≡ 5 mol of C 2 O 42-( KHC 2 O 4) (100 × 0.02/1000) mol MnO 4-=0.005 mol KHC 2 O 4

Q. Moles of $KH C_{2} O_{4}$ (potassium acid oxalate) required to reduce $100 m L$ of $0.02 M K M n O_{4}$ in acidic medium $($ to $M n^{2 +})$ is

$0.002$

$0.005$

$0.001$

$0.007$

$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} \to M n^{2 +} + 10 C O_{2}$
$2 m o l$ of $M n O_{4}^{-} \equiv 5 m o l$ of $C_{2} O_{4}^{2 -} (KH C_{2} O_{4})$
$\frac{100 \times 0.02}{1000} m o l M n O_{4}^{-} = 0.005 m o l KH C_{2} O_{4}$

Q. Moles of KHC2​O4​ (potassium acid oxalate) required to reduce 100mL of 0.02MKMnO4​ in acidic medium ( to Mn2+) is

0.002

0.005

0.001

0.007