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

Copper (II) chloride.

Step-by-step 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*(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*(1molCuCl_2)/(134.45gCuCl_2)*(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.