)a chemist combined 0.786 g of aluminum with 25.0 ml of a 0.21 m solution of cucl2. which substances will be present in the flask after the reaction has completed?

a) solid copper, solid aluminum, and aqueous aluminum chloride
b) solid copper and aqueous aluminum chloride
c) solid aluminum and aqueous copper(ii) chloride
d) solid copper and a solution containing cu2+ (aq), al3+ (aq), and cl− (aq)
e) solid aluminum and a solution containing cu2+ (aq), al3+ (aq), and cl− (aq)

In the flask solid copper, solid aluminum, and aqueous aluminum chloride will be present after the reaction

Explanation:

Step 1: Data given

Mass of aluminium = 0.786 grams

Volume of 0.21 M CuCl2 solution = 25.0 mL = 0.025 L

Molar mass of Al = 26.98 g/mol

Step 2: The balanced equation

CuCl2 + Al → AlCl2 + Cu

Step 3: Calculate moles Al

Moles Al = mass Al / molar mass Al

Moles Al = 0.786 grams / 26.98 g/mol

Moles Al = 0.0291 moles

Step 4: Calculate moles CuCl2

Moles CuCl2 = molarity * volume

Moles CuCl2 = 0.21 M * 0.025 L

Moles CuCl2 = 0.00525 moles

Step 5: Calculate the limiting reactant

For 1 mol CuCl2 we need 1 mol Al to produce 1 mol AlCl2 and 1 mol Cu

CuCl2 is the limiting reactant, it will completely be consumed ( 0.00525 moles).

Al is in excess. There will react 0.00525 moles. There will remain 0.0291 - 0.00525 = 0.02385 moles Al

Step 6: Calculate moles of products

There will be produce 0.00525 moles of AlCl2 and 0.00525 moles of Cu

This means in the flask solid copper, solid aluminum, and aqueous aluminum chloride will be present after the reaction

the correct answer is:

a. formation of precipitation