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. A single replacement reaction takes place. What best explains the state of the reaction mixture after the reaction? Unbalanced equation: CuCl2 + Al → AlCl3 + Cu (4 points)

Group of answer choices

Less than 6.0 grams of copper is formed, and some aluminum is left in the reaction mixture.

More than 6.5 grams of copper is formed, and some aluminum is left in the reaction mixture.

Less than 6.0 grams of copper is formed, and some copper chloride is left in the reaction mixture.

More than 6.5 grams of copper is formed, and some copper chloride is left in the reaction mixture.

Less than 6.0 grams of copper is formed, and some aluminum is left in the reaction mixture.

Explanation:

Based on the balanced equation of reaction and the reacting masses, approximately 0.36 grams of Cu are produced because copper (II) chloride acts as a limiting reactant.

What is the amount of copper produced from the reaction?

The amount of copper produced is determined from the balanced equation of the reaction.

The equation of the reaction is given below:

3 CuCl₂ + 2 Al → 2 AlCl₃ + 3 Cu

molar mass of CuCl₂ = 132 g/mol

molar mass of Al = 27.0 g

molar mass of Cu = 64.0 g

moles of aluminium in 0.50 g = 0.50 / 27 = 0.018 moles

moles of CuCl₂ in 0.75 g = 0.75/132 = 0.00568 moles

Therefore, CuCl₂ is the limiting reactant

3 moles of CuCl₂ produces 3 moles of Cu

0.00568 moles of CuCl₂ will produce 0.00568 moles of Cu

Mass of 0.005  moles of Cu = 0.00568 * 64 = 0.36 g of Cu

Therefore, approximately 0.36 grams of Cu are produced because copper (II) chloride acts as a limiting reactant.

Learn more about limiting reactant at:

answer: your mm

explanation:

the answer to this question is 3.