Use Fe = 56, Na = 23, S = 32, C = 12, O = 16. A 4.38 g mixture of Na2C2O4 and FeSO4 (no other reductant) consumed 200 mL of 0.04 M KMnO4 in acid (to Mn^2+). The mass of FeSO4 in the mixture is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx145

Use Fe = 56, Na = 23, S = 32, C = 12, O = 16. A $4.38 g$ mixture of $Na X_{2} C X_{2} O X_{4}$ and $FeSO X_{4}$ (no other reductant) consumed $200 mL$ of $0.04 M$ $KMnO X_{4}$ in acid (to $Mn X^{2 +}$ ). The mass of $FeSO X_{4}$ in the mixture is:

Let

a

= mol oxalate (

n = 2

) and

b

= mol

FeSO X_{4}

(

n = 1

). Equivalents:

2 a + b = 0.200 \times 0.04 \times 5 = 0.04

. Mass:

134 a + 152 b = 4.38

. Solving gives

a = 0.010

b = 0.020

. Mass of

FeSO X_{4}

= 0.020 \times 152 = 3.04 g