For the reaction

? Al + ? CuSO4 ↽⇀? Al2(SO4)3+? Cu
a maximum of how many moles of Al2(SO4)3
could be formed from 5.49 mol of Al and
4.09 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 identify the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thereby limiting the amount of product that can be formed.

To find the limiting reactant, we compare the moles of each reactant with the stoichiometric ratios provided by the balanced chemical equation.

First, we need to balance the chemical equation for the reaction:
2 Al + 3 CuSO4 ⟶ Al2(SO4)3 + 3 Cu

The balanced equation shows that 2 moles of Al react with 3 moles of CuSO4 to produce 1 mole of Al2(SO4)3 and 3 moles of Cu.

Next, we compare the moles of Al and CuSO4 to determine the limiting reactant.

For Al: 5.49 mol
For CuSO4: 4.09 mol

Using the stoichiometric ratio from the balanced equation, we can calculate the theoretical number of moles of Al2(SO4)3 that would be produced if either reactant were to be completely consumed.

If we assume that Al is completely consumed:
5.49 mol Al × (1 mol Al2(SO4)3 / 2 mol Al) = 2.745 mol Al2(SO4)3

If we assume that CuSO4 is completely consumed:
4.09 mol CuSO4 × (1 mol Al2(SO4)3 / 3 mol CuSO4) = 1.363 mol Al2(SO4)3

Comparing these values, we can see that 1.363 mol Al2(SO4)3 is the smaller value. This means that CuSO4 is the limiting reactant, and 1.363 mol is the maximum number of moles of Al2(SO4)3 that can be formed.

Therefore, the maximum number of moles of Al2(SO4)3 that could be formed from 5.49 mol of Al and 4.09 mol of CuSO4 is 1.363 mol.