For the reaction
? Al + ? CuSO4 ↽⇀? Al2(SO4)3+? Cu
a maximum of how many moles of Al2(SO4)3
could be formed from 5.95 mol of Al and
1.31 mol of CuSO4?
Answer in units of mol.
To determine the maximum number of moles of Al2(SO4)3 that can be formed, we need to use the concept of limiting reactants. The reactant that will be completely consumed first will limit the amount of product formed.
First, let's determine the balanced chemical equation for the reaction:
2Al + 3CuSO4 ⟶ Al2(SO4)3 + 3Cu
From the balanced chemical equation, we can see that the ratio of Al to Al2(SO4)3 is 2:1. This means that 2 moles of Al will produce 1 mole of Al2(SO4)3.
Similarly, the ratio of CuSO4 to Al2(SO4)3 is 3:1. This means that 3 moles of CuSO4 will produce 1 mole of Al2(SO4)3.
Now, let's calculate the number of moles of Al2(SO4)3 that can be formed using the given amounts of Al and CuSO4:
For Al:
Given moles of Al = 5.95 mol
Moles of Al2(SO4)3 = 5.95 mol Al × (1 mol Al2(SO4)3 / 2 mol Al) = 2.98 mol Al2(SO4)3
For CuSO4:
Given moles of CuSO4 = 1.31 mol
Moles of Al2(SO4)3 = 1.31 mol CuSO4 × (1 mol Al2(SO4)3 / 3 mol CuSO4) = 0.44 mol Al2(SO4)3
The limiting reactant is CuSO4 because it produces a smaller amount of Al2(SO4)3. Therefore, the maximum moles of Al2(SO4)3 that can be formed is 0.44 mol.
To determine the maximum number of moles of Al2(SO4)3 that could be formed, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thus limiting the amount of product that can be formed.
To find the limiting reactant, we compare the number of moles of each reactant with the stoichiometric ratio from the balanced chemical equation. Let's balance the equation first:
2 Al + 3 CuSO4 ↽⇀ Al2(SO4)3 + 3 Cu
From the balanced equation, the stoichiometric ratio is 2 moles of Al to 1 mole of Al2(SO4)3. Similarly, the stoichiometric ratio is 3 moles of CuSO4 to 1 mole of Al2(SO4)3.
Now, let's calculate the number of moles of Al2(SO4)3 that could be formed from each reactant:
For Al:
Using the stoichiometric ratio of 2 moles of Al to 1 mole of Al2(SO4)3:
Number of moles of Al2(SO4)3 = (5.95 mol Al) × (1 mol Al2(SO4)3 / 2 mol Al) = 2.98 mol Al2(SO4)3
For CuSO4:
Using the stoichiometric ratio of 3 moles of CuSO4 to 1 mole of Al2(SO4)3:
Number of moles of Al2(SO4)3 = (1.31 mol CuSO4) × (1 mol Al2(SO4)3 / 3 mol CuSO4) = 0.437 mol Al2(SO4)3
Comparing the calculated number of moles of Al2(SO4)3 for each reactant, we can see that the limiting reactant is CuSO4 because it results in the smallest number of moles of Al2(SO4)3.
Therefore, the maximum number of moles of Al2(SO4)3 that could be formed from 5.95 mol of Al and 1.31 mol of CuSO4 is 0.437 mol.
Matt, this is worked almost the same way as your posts under CJ and Lacey EXCEPT that this is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
Here is an worked example of a limiting reagent problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html