Chemistry
Calculate the percentage of iron in K3Fe(CN)6. [K = 39, Fe = 56, C = 12, N = 14]
Stoichiometry
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Answer
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
ā“ 100 g of K3Fe(CN)6 contains x 100 = 17.02% of iron.
Hence, percentage by weight of iron in K3Fe(CN)6 is 17.02%
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