Consider the following reaction.

2Fe2O3(s)+3C(s)−→−heat4Fe(s)+3CO2(g)

Calculate the number of grams of Fe2O3 needed to react with 13.0 g C.

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Answer:

Number of moles of C = 13.0 g / 12.0107 g/mol = 1.08 mole

From the balanced equation we can say that

3 mole of C requires 2 mole of Fe2O3 so

1.08 mole of C will require

= 1.08 mole of C *(2 mole of Fe2O3 / 3 mole of C)

= 0.720 mole of Fe2O3

mass of 1 mole of Fe2O3 = 159.69 g so

the mass of 0.720 mole of Fe2O3 = 115 g

Therefore, the mass of Fe2O3 produced would be 115 g

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