Question

A chemist wants to extract copper metal from copper chloride solution. The chemist places 1.50 grams of aluminum foil in a solution of 14 grams of copper (II) chloride. (single replacement reaction.) What best explains the state of the reaction mixture afterward​?
a. Less than 6.0 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.
b. More than 6.5 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.
c. Less than 6.0 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.
d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Answer 1

Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

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

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.

Thus 54 g of of aluminium react with 270 g of copper chloride.

1.50 g of aluminium react with=$\frac{270}{54}\times 1.50=7.5 g$of copper chloride.

3 moles of copper chloride gives 3 moles of copper.

7.5 g of copper chloride gives 7.5 g of copper.

Answer 2

Answer: The correct answer is Option c.

Explanation:

For the reaction of aluminium and copper (II) chloride, the equation follows:

[ex]2Al+3CuCl_3\rightarrow 2AlCl_3+3Cu[/tex]

The molarity is calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    ....(1)

• For Aluminium:

Given mass = 1.5 g

Molar mass = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{1.5g}{27g/mol}=0.055mol$

• For Copper chloride:

Given mass = 14g

Molar mass = 134.45 g/mol

Putting values in equation 1, we get:

$\text{Moles of copper chloride}=\frac{14g}{134.45g/mol}=0.104mol$

• By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of copper chloride

So, 0.055 moles of aluminium will react with = $\frac{3}{2}\times 0.055=0.0825mol$ of copper chloride.

As, the required amount of copper chloride is less than the given amount. Hence, it is present in excess.

Therefore, Aluminium is considered as a limiting reagent because it limits the formation of product.

• By Stoichiometry of the above reaction:

2 moles of aluminium produces 3 moles of copper metal

So, 0.055 moles of aluminium will produce = $\frac{3}{2}\times 0.055=0.0825mol$ of copper metal.

• To calculate the mass of copper produced, we use equation 1:

Number of moles of copper = 0.0825 mol

Molar mass of copper = 63.5 g/mol

Putting values in the equation 1, we get:

$0.0825mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=5.238g$

As, aluminium is fully utilized in the reaction. So, some copper chloride is left in the reaction mixture.

Hence, the correct answer is Option c.