Use K = 39, Cr = 52, O = 16, Fe = 56, S = 32. In acidic solution, 0.49 g of K2Cr2O7 (reduced to Cr^3+) oxidises what mass of FeSO4 (to Fe^3+)?

JEE Main levelSingle correct optionDifficulty 2 of 3rx125

Use K = 39, Cr = 52, O = 16, Fe = 56, S = 32. In acidic solution, $0.49 g$ of $K X_{2} Cr X_{2} O X_{7}$ (reduced to $Cr X^{3 +}$ ) oxidises what mass of $FeSO X_{4}$ (to $Fe X^{3 +}$ )?

M (K X_{2} Cr X_{2} O X_{7}) = 294

, so moles

= 0.49/294 = 1.667 \times 1 0^{- 3}

. Equivalents

= 6 \times

moles

= 0.01

. Equivalents of

FeSO X_{4}

equal 0.01, and

M (FeSO X_{4}) = 152

, giving

0.01 \times 152 = 1.52 g