Answer:
Copper (II) chloride.
Explanation:
Hello,
In this case, considering the described reaction which is also given as:
[tex]2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu[/tex]
For us to identify the limiting reactant we first compute the available moles of aluminium:
[tex]n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl[/tex]
Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:
[tex]n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2} =5.69molAl[/tex]
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.
