Achemist wants to extract copper metal from copper chloride solution. the chemist places 0.50 grams of aluminum foil in a solution containing 0.75 grams of copper (ii) chloride. a single replacement reaction takes place. (ii) chloride. a single replacement reaction takes place. which statement explains the maximum amount of copper that the chemist can extract using this reaction? a) approximately 0.36 grams, because copper (ii) chloride acts as a limiting reactant b) approximately 1.8 grams, because copper (ii) chloride acts as a limiting reactant c) approximately 0.36 grams, because aluminum acts as a limiting reactant d) approximately 1.8 grams, because aluminum acts as a limiting reactant

Answer & Explanation:

The replacement of aluminium (Al) by copper (Cu) can be represented by the following chemical equation:

3CuCl2 + 2Al  = 2AlCl3 + 3Cu

We are given 0.50g of Al, and 0.75g of CuCl2

We calculate the maximum amount of copper by assuming one of the reactants (CuCl2 or Al) will be depleted.

Calculations:

The corresponding molecular masses are :

3CuCl2 = 3 (63.546+2*35.457) = 403.38 (1)

2Al = 2*26.9815 = 53.963(2)

2AlCl3 = 2 (26.9815+3*35.457) = 266.705  (we don't need to know this!)

3Cu = 3*63.546 = 190.638(3)

Assuming 0.50g of aluminium will be depleted with unlimited CuCl2, the amount of copper recuperated will be:

M1 = 0.50 *( (3)/(2) ) = 0.50*190.638/53.963 = 1.766 g

If 0.75g of CuCl2 were depleted with unlimited Al, the amount of copper recuperated will be

M2 = 0.75 * ( (3) / (1) ) = 0.75 * 190.638 / 403.38 = 0.354 g  

Therefore CuCl2 is the limiting reagent.

Since we can obtain the maximum amount of copper from the limiting reagent (CuCl2), so the required maximum amount is 0.35 g of copper.

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³  mol.From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³  mol of Al foil reacts completely with 5.578 x 10⁻³  mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of  CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³  mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

So, the answer is:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.