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What is the limiting reactant when 14.7 grams of aluminum reacts with 13.6 grams of CuCl2? What is the theoretical yield of copper?

2Al(s)+3CuCl2(aq) --> 2AlCl3(aq)+3Cu(s)​

User Ajo Koshy
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Answer:To determine the limiting reactant, we need to compare the amounts of aluminum and CuCl2 to see which one will run out first.

First, we need to convert the given masses to moles. The molar masses of Al and CuCl2 are:

Al: 26.98 g/mol

CuCl2: Cu (63.55 g/mol) + 2 Cl (35.45 g/mol) = 134.45 g/mol

Using these molar masses, we can calculate the number of moles of each reactant:

moles of Al = 14.7 g / 26.98 g/mol = 0.545 moles

moles of CuCl2 = 13.6 g / 134.45 g/mol = 0.101 moles

Now we can use the stoichiometry of the balanced chemical equation to determine the amount of product that each reactant can produce. We will compare the theoretical yield of copper from each reactant to determine the limiting reactant.

From the balanced equation, we can see that 2 moles of Al react with 3 moles of CuCl2 to produce 3 moles of Cu. Therefore, the theoretical yield of copper from the 0.545 moles of Al is:

theoretical yield of Cu from Al = (0.545 mol Al) x (3 mol Cu / 2 mol Al) x (63.55 g Cu / mol Cu) = 54.9 g Cu

Similarly, the theoretical yield of copper from the 0.101 moles of CuCl2 is:

theoretical yield of Cu from CuCl2 = (0.101 mol CuCl2) x (3 mol Cu / 3 mol CuCl2) x (63.55 g Cu / mol Cu) = 6.44 g Cu

Since the theoretical yield of Cu from CuCl2 is lower than the theoretical yield of Cu from Al, CuCl2 is the limiting reactant.

Therefore, the theoretical yield of copper is 6.44 g.

Step-by-step explanation:

User Tim Malseed
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