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2Al(s) + 3CuSO4(aq) --> Al2(SO4)3(aq) + 3Cu(s) 2. According to the reaction in question 1 above, how many grams of solid copper will theoretically be produced when 2.2 g of aluminum are reacted with 14.4 g of copper(II) sulfate? Which reactant is the limiting reactant? Show your work for this problem

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Answer:

5.73 g of solid copper, can be  theoretically produced in this reaction.

Explanation:

We start from the begining with the balanced equation:

2Al (s) + 3CuSO₄ (aq) → Al₂(SO₄)₃ (aq)  +  3Cu (s)

To determine the limiting reactant we need to know the moles of each.

2.2 g / 26.98 g/mol = 0.0815 moles of Al

14.4 g / 159.6 g/mol = 0.0902 moles of copper (II) sulfate.

Ratio is 2:3. Let's find out the limtiing reactant,

2 moles of Al react to 3 moles of copper (II) sulfate

Then 0.0815 moles of Al, may react to (0.0815 . 3) / 2 = 0.122 moles of salt.

We only have 0.0902 moles of sulfate and we need 0.122 moles to complete the reaction, as the sulfate we have, is not enough, the salt is the limting reagent. We use this to determine the solid copper that can be theoretically produced.

This ratio is 3:3, so If we have 0.0902 moles of copper (II) sulfate, we may produce 0.0902 moles of solid copper.

Let's conver the moles to mass → 0.0902 mol . 63.5 g/mol = 5.73 g

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