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Iron(III) oxide reacts with carbon monoxide according to the equation: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) A reaction mixture initially contains...

Iron(III) oxide reacts with carbon monoxide according to the equation: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) A reaction mixture initially contains 22.95 g Fe2O3 and 14.26 g CO.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

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Answer #1

The number of mols of Fe2O3 = 22.95 g / 159.69 g/mol = 0.1437 mol

The number of mols of CO = 14.26 g / 28 g/mol = 0.509 mol

As per the balanced equation, 1 mol of Fe2O3 requires 3 mol CO. Therefore, 0.1437 mol Fe2O3 requires (0.1437 mol)3 = 0.4311 mol

But in the mixture its around 0.509 mols of CO is there, hence we conclude that CO is in excess

The number of mols oCO left = 0.509 mol - 0.4311 mol = 0.0779 mol

Mass of excess reactant = 0.0779 mol (28 g/mol) = 2.18 g

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