# 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

#### Answer:

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

#### Explanation:

Start with a balanced equation.

${\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.