0.6250 g of an iron ore is dissolved in acid and all the iron is reduced to Fe^2+ (excess reductant removed). The solution needs 28.4 mL of 0.0200 M K2Cr2O7. Taking Fe=56, the percentage of iron in the ore is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq198

$0.6250 g$ of an iron ore is dissolved in acid and all the iron is reduced to $Fe X^{2 +}$ (excess reductant removed). The solution needs $28.4 mL$ of $0.0200 M$ $K X_{2} Cr X_{2} O X_{7}$ . Taking $Fe = 56$ , the percentage of iron in the ore is:

Cr X_{2} O X_{7} X^{2 -}

takes 6 electrons, so mmol of

Fe X^{2 +}

= 28.4 \times 0.0200 \times 6 = 3.408

. Mass of Fe

= 3.408 \times 56 = 190.8

mg. Percentage

= 190.8/625.0 \times 100 = 30.54%