Question

10.0 grams of copper sulfate are combined with 10.0 grams of aluminum.

3 CuSO4 + 2 Al --> Al2(SO4)3 + 3 Cu
What is the limiting reactant? (just pick 1 product to convert each reactant to- just make sure the product converted to is the same for both reactants.)

Answer 1

To determine the limiting reactant, we first need to calculate how many moles of each reactant we have. We can use the molar mass of each reactant to convert the given mass to moles.

For copper sulfate:

10.0 g CuSO4 / 159.61 g/mol = 0.0628 moles

For aluminum:

10.0 g Al / 26.98 g/mol = 0.371 moles

Next, we need to use the coefficients from the balanced chemical equation to convert the moles of each reactant to the moles of one of the products. In this case, we will convert the moles of each reactant to moles of Al2(SO4)3.

From the balanced equation:

3 moles CuSO4 = 1 mole Al2(SO4)3

2 moles Al = 1 mole Al2(SO4)3

Therefore, to find the moles of Al2(SO4)3 that can be produced from 0.0628 moles of CuSO4 and 0.371 moles of Al, we divide each value by the respective coefficients from the balanced equation:

0.0628 moles CuSO4 / 3 = 0.0210 moles Al2(SO4)3

0.371 moles Al / 2 = 0.185 moles Al2(SO4)3

Here, 0.185 moles Al2(SO4)3 is the limiting reactant as it is the lowest amount of one of the products that can be formed by converting both reactants. Therefore, the limiting reactant is Aluminum.