дано

m(C) = 2.25 g

m практ (Fe) = 11.2 g

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w(Fe) -?

2Fe2O3+3C-->4Fe+3CO2

M(C) = 12 g/mol

n(C) = m/M = 2.25 / 12 = 0.1875 mol

3n(C) = 4n(Fe)

n(Fe) = 4*0.1875 / 3 = 0.25 mol

M(Fe) = 56 g/mol

m теор (Fe) = n(Fe)*M(Fe) = 0.25 * 56 = 14 g

W(Fe) = m практ (Fe) / m теор (Fe) * 100% = 11.2 / 14 * 100% = 80 %

ответ 80%