To balance the chemical equation for the rusting of iron which forms iron(III) oxide (Fe₂O₃), you will need to find the correct coefficients.
The unbalanced equation is:
\( \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 \)
To balance it:
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There are 2 iron (Fe) atoms in iron(III) oxide (Fe₂O₃), so you need 2 Fe on the left:
\( 2 \text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 \)
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In Fe₂O₃, there are 3 oxygen (O) atoms. Since O₂ is diatomic (contains 2 O atoms), you will need 1.5 O₂ to get 3 O atoms. However, to avoid using a fractional coefficient, we can multiply the entire equation by 2:
This gives:
\( 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \)
So the balanced coefficients are:
4 for Fe, 3 for O₂, and 2 for Fe₂O₃.
Therefore, the balanced equation is:
\( \mathbf{4} \text{Fe} + \mathbf{3} \text{O}_2 \rightarrow \mathbf{2} \text{Fe}_2\text{O}_3 \)
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