# If 1.35 g of aluminum is mixed with 6.0 moles of copper (II) chloride, what mass of copper would be produced?

Aug 22, 2017

Since aluminum is the limiting reactant, so the reaction will produce $\text{4.77 g Cu}$.

#### Explanation:

${\text{2Al + 3CuCl}}_{2}$$\rightarrow$$\text{2AlCl"_3 + "3Cu}$

Convert $\text{1.35 g Al}$ to moles by dividing by its molar mass, $\text{26.982 g/mol}$. Since molar mass is a fraction, g/mol, you can divide by multiplying the given mass by the inverse of the molar mass, mol/g .

1.35color(red)cancel(color(black)("g Al"))xx(1"mol Al")/(26.982color(red)cancel(color(black)("g Al")))="0.0500 mol Al"

You need to determine the limiting reactant (or limiting reagent). You need to determine which element or compound produces the least amount of $\text{Cu}$.

The process involves multiplying the mol $\text{Al}$ and $\text{CuCl"_2}$ by the mole ratio between each of them and $\text{Cu}$, and then multiply by the molar mass of $\text{Cu}$, $\text{63.546 g/mol}$.

Possible Mass of Cu from Al

0.0500color(red)cancel(color(black)("mol Al"))xx(3color(red)cancel(color(black)("mol Cu")))/(2color(red)cancel(color(black)("mol Al")))xx(53.546"Cu")/(1color(red)cancel(color(black)("mol Cu")))="4.77 g Cu"

Possible Mass of Cu from ${\text{CuCl}}_{2}$

6.0color(red)cancel(color(black)("g CuCl"_2))xx(3color(red)cancel(color(black)("mol Cu")))/(3color(red)cancel(color(black)("mol CuCl"_2)))xx(63.546"g Cu")/(1color(red)cancel(color(black)("mol Cu")))="380 g Cu"

Under the conditions given, Al is the limiting reactant because it will produce only $\text{4.77 g Cu}$, whereas, if there were enough Al present, the $\text{CuCl"_2}$ would produce $\text{380 g Cu}$. However, under the conditions given, the Al would run out before the reaction could produce $\text{380 g Cu}$. The ${\text{CuCl}}_{2}$ is the reactant in excess.