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PtichkaEL [24]
3 years ago
9

Al (s) + Fe2O3 (s) --> Al2O3 (s) + Fe (s) (needs balancing)

Chemistry
1 answer:
Andreas93 [3]3 years ago
4 0

Answer: 134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of Fe_2O_3.

Solution:

Given :Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+Fe(s)

After balancing, 2Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+2Fe(s)

As we are given Fe_2O_3 is in excess. Hence, Al will be considered as limiting reagent because it limits the formation of product.

\text{Number of moles of}Al=\frac{\text{mass of the compound}}{\text{Molecular mass of the compound}}=\frac{65.2 g}{26.98 g/mol}=2.4166 moles

According to reaction ,2 moles of Al reacts with one mole of Fe_2O_3 to give 2 moles of Fe then 2.4166 moles of Al will give :

=\frac{2}{2}\times 2.4166\text{moles of}Fe= 2.4166 moles

Mass of Fe produced =\text{Number of moles of compound}\times \text{Molecular mass of the compound}=2.4166 moles\times 55.85g/mol=134.967 g

134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of Fe_2O_3.

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