Question

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solution of 0.40 grams of copper (II) chloride. A single replacement reaction takes place. What are the likely observations when the reaction stops?
About 0.90 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.
About 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.
About 0.90 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.
About 0.20 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Answer 1

Answer: The correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

Explanation:

To calculate the number of moles, we use the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

• Moles of Aluminium:

Molar mass of aluminium = 27 g/mol

Given mass of aluminium =  0.25g

Putting values in above equation, we get:

$Moles=\frac{0.25g}{27g/mol}=0.00925mol$

• Moles of Copper chloride:

Molar mass of copper chloride = 134.45 g/mol

Given mass of copper chloride=  0.40g

Putting values in above equation, we get:

$Moles=\frac{0.40g}{134.45g/mol}=0.00297mol$

For the given chemical equation:  

$4Al(s)+3CuCl_2(aq.)\rightarrow 2Al_2Cl_3(aq.)++3Cu(s)$

By Stoichiometry,

3 moles of Copper chloride reacts with 4 moles of aluminium

So, 0.00297 moles of copper chloride will react with = $\frac{4}{3}\times 0.00297$ = 0.00396 moles of Aluminium.

As, the required moles of aluminium is less than the given moles of aluminium. Hence, it is considered as the excess reagent.

Copper chloride is the limiting reagent in the given chemical reaction because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of copper chloride produces 3 moles of copper metal.

So, 0.00297 moles of copper chloride will produce = $\frac{3}{3}\times 0.00297$ = 0.00297 moles of copper metal.

Now, to calculate the given mass of copper metal, we use the equation required to calculate moles:

Molar mass of copper = 63.5 g/mol

Putting values in that equation, we get:

$0.00297mol=\frac{\text{given mass}}{63.5g/mol}$

Given mas of copper metal = 0.188grams (approx 0.20grams)

For the given reaction, some of the aluminum is left behind because it is the excess reagent in the reaction.

Hence, the correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

Answer 2

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solution of 0.40 grams of copper (II) chloride. A single replacement reaction takes place. What are the likely observations when the reaction stops?

About 0.90 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

About 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture. CORRECT ANSWER

About 0.90 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

About 0.20 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.