Calculate the number of moles of Al2O3 that are produced when 0.6 mol of Fe is produced in the following reaction. 2Al(s) + 3Fe(s)-- 3Fe(s) + Al2O3(s)
To calculate the number of moles of Al2O3 produced, you need to use the stoichiometry of the balanced chemical equation.
The stoichiometry of the equation tells us that for every 2 moles of Al, we will produce 1 mole of Al2O3. Therefore, the mole ratio between Al2O3 and Al is 1:2.
Since we are given the number of moles of Fe produced (0.6 mol), we can use this information to find the number of moles of Al2O3 produced.
First, we need to determine the moles of Al needed to produce 0.6 mol of Fe. From the balanced equation, we know that the mole ratio between Al and Fe is 3:2. Therefore, we can set up a proportion to find the moles of Al:
(0.6 mol Fe / 3 mol Fe) = (x mol Al / 2 mol Al)
Cross multiplying, we have:
2 mol Al * 0.6 mol Fe = 3 mol Fe * x mol Al
1.2 mol Al = 3x mol Al
Dividing both sides by 3, we find that x (the moles of Al) is 0.4 mol.
Now that we know the moles of Al, we can use the mole ratio between Al and Al2O3 to find the moles of Al2O3:
0.4 mol Al * (1 mol Al2O3 / 2 mol Al) = 0.2 mol Al2O3
Therefore, 0.2 moles of Al2O3 are produced when 0.6 moles of Fe are produced in the given reaction.