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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Question
Question
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