How much iron can be extracted from 354 g of ferrum (bivalent) chloride FeCl2?

univerkov | December 3, 2020 | education

n (FeCl2) = m \ M = 354 \ (56 + 35.5 * 2) = 2.79 mol
n (Fe) = n (FeCl2) = 2.79 mol
m (Fe) = n * M = 2.79 * 56 = 156.24 g
Answer: m (Fe) that can be mined = 156.24 g

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