Question

If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu

Answer 1

Answer:

Copper (II) chloride.

Explanation:

Hello,

In this case, considering the described reaction which is also given as:

$2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu$

For us to identify the limiting reactant we first compute the available moles of aluminium:

$n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl$

Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:

$n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2} =5.69molAl$

Thereby, we can infer aluminium is in excess since less moles are consumed than available whereas the copper (II) chloride is the limiting reactant.

Best regards.