Fe2O3+2Al-->2Fe+Al2O3 how many grams of Fe2O3 are needed to produce .905 mol ofA

Question

Fe2O3+2Al-->2Fe+Al2O3
how many grams of Fe2O3 are needed to produce .905 mol ofAl2O3 Fe2O3+2Al-->2Fe+Al2O3
how many grams of Fe2O3 are needed to produce .905 mol ofAl2O3

Explanation / Answer

Fe2O3 +2Al -->2Fe + Al2O3 From the equation, we know that 1 mole ofFe2O3 reacts to produce 1 mole ofAl2O3. Therefore:     moles Fe2O3 = molesAl2O3 = 0.905 mol Now multiply moles by molar mass to get the grams ofFe2O3 required:     molar mass = 55.845 * 2 + 15.999 * 3 = 159.687g.mol-1     mass = moles * molar mass             = 0.905 mol * 159.687 g.mol-1             = 144.5 g

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