Aluminium reacts with iron (III) oxide, Fe₂O₃, to give iron metal and aluminium oxide, Al₂O₃.
The a) part of the question was to find the balanced equation of this reaction - which is:
2Al + Fe₂O₃ = 2Fe + Al₂O₃
b) In an experiment, 32.0g of iron (III) oxide was reacted with 16.2g of aluminium. Which of the two reactants is the limiting reactor?
Ans: In the reaction, the mole ratio of Al to Fe₂O₃ is 2:1, so 0.600 mol of Al could react with 0.300 mol of Fe₂O₃. There is only 0.200 mol available, though, so the Fe₂O₃ is the limiting reactant and Al is in excess.
So, what is the maximum mass of iron that could be collected at the end of this experiment?
Calculate the maximum mass of iron that could be collected at the end of this experiment?
2017-09-30 1:12 am
回答 (2)
2017-09-30 2:12 am
✔ 最佳答案
2Al + Fe₂O₃ = 2Fe + Al₂O₃ According to the coefficients in this equation, two moles of aluminum reacts with one mole of iron (III) oxide to produce two moles of iron and one mole of aluminum oxide. Let’s determine the number of moles of Aluminum and iron (III) oxide.
For Al, n = 16.2 ÷17 = 0.6
Let’s determine the mass of one mole of Fe₂O₃.
Mass = 55.847 * 2 + 3 * 16 = 159.694
n = 32 ÷ 32 ÷ 159.64
This is approximately 0.2 mole. The ratio of these two numbers is approximately 3:1. The ratio is supposed to be 2:1. This means there excess aluminum. This means iron (III) oxide is the limiting reactant. All of iron in the iron (III) oxide becomes iron.
Mass of Fe₂ = 2 * 55.8 = 111.6 grams
Fraction = 111.6 ÷ 159.694
This is approximately 0.7. To determine the mass of iron that is produced, multiply the mass of iron (III) oxide by the fraction.
Mass = 32 * 111.6 ÷ 159.694
This is approximately 22 grams of iron. I hope the method that I used to solve this problem is helpful for you!
2017-09-30 1:25 am
(32.0 g Fe2O3) / (159.6882 g Fe2O3/mol) = 0.20039 mol Fe2O3
(16.2 g Al) / (26.98154 g Al/mol) = 0.60041 mol Al
0.20039 mole of Fe2O3 would react completely with 0.20039 x (2/1) = 0.40078 mole of Al, but there is more Al present than that, so Al is in excess and Fe2O3 is the limiting reactant.
(0.20039 mol Fe2O3) x (2 mol Fe / 1 mol Fe2O3) x (55.8450 g Fe/mol) = 22.4 g Fe
(16.2 g Al) / (26.98154 g Al/mol) = 0.60041 mol Al
0.20039 mole of Fe2O3 would react completely with 0.20039 x (2/1) = 0.40078 mole of Al, but there is more Al present than that, so Al is in excess and Fe2O3 is the limiting reactant.
(0.20039 mol Fe2O3) x (2 mol Fe / 1 mol Fe2O3) x (55.8450 g Fe/mol) = 22.4 g Fe
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