Molecular weight of K3Fe(CN)6 = 3(39) + 56 + 6(12 + 14)
= 117 + 56 + 156
= 329 g
329 g of K3Fe(CN)6 contains 56 g of iron
∴ 100 g of K3Fe(CN)6 contains 56\over329 × 100 = 17.02% of iron.
Hence, percentage by weight of iron in K3Fe(CN)6 is 17.02%
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