дано

m(Fe3O4) = 500 g

η(Fe) = 85%

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m практ (Fe)-?

3Fe3O4+8Al-->9Fe+4Al2O3

M(Fe3O4) = 232 g/mol

n(Fe3O4) = m/M = 500  / 232 = 2.16 mol

3n(Fe3O4) = 9n(Fe)

n(Fe) = 9*2.16 / 3 = 6.48 mol

M(Fe) = 56 g/mol

m теор(Fe) = n*M =6.48 * 56 = 362.88 g

m практ(Fe) = m теор(Fe) * η(Fe) /100% = 362.88 * 85% / 100% = 308.448 g

ответ 308.448 г