Examine this equation and answer the questions that follow: 2Al + Fe2O3 → Al2O3 + 2Fe How many grams of Fe can be produced when 10.0g of Al is reacted with an excess (unlimited) supply of Fe2O3?

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Examine this equation and answer the questions that follow: 2Al + Fe2O3 → Al2O3 + 2Fe How many grams of Fe can be produced when 10.0g of Al is reacted with an excess (unlimited) supply of Fe2O3?

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Answer 1 · 2018-02-21T16:08:39

The molar ratio of $A l \div F e$ $AldivFe$ will be $2 \div 2 = 1 \div 1$ $2div2=1div1$

Explanation:

Since the atomic mass ratio is $m (A l) \div m (F e) = 26.98 \div 55.85$ $m(Al)divm(Fe)=26.98div55.85$ :

$\frac{10.0}{26.98} = \frac{m (F e)}{55.85} \to m (F e) = \frac{10.0 \times 55.85}{26.98} \approx 20.7 g$ $10.0/26.98=(m(Fe))/55.85->m(Fe)=(10.0xx55.85)/26.98~~20.7g$

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Examine this equation and answer the questions that follow: 2Al + Fe2O3 → Al2O3 + 2Fe How many grams of Fe can be produced when 10.0g of Al is reacted with an excess (unlimited) supply of Fe2O3?

1 Answer

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