Question

What is the limiting reactant when 415 mL of 3.0M copper(II) chloride solution is added to 25.0 g of solid aluminum?

Answer 1

Answer:

The answer to your question is Copper (II) chloride

Explanation:

Data

Limiting reactant = ?

volume of CuCl₂ = 415 ml

[CuCl₂] = 3.0 M

mass of Al = 25 g

Process

1.- Calculate the mass of CuCl₂ in solution

Molarity = moles / volume

moles = Molarity x volume

moles = 3 x 0.415

moles = 1.245

Molar mass of CuCl₂ = 63.5 + (35.5 x 2)

                                  = 63.5 + 71

                                  = 134.5 g

                   134.5 g ----------------------- 1 mol

                      x        ----------------------- 1.245 moles

                      x = (1.245 x 134.5)/1

                      x = 167.5 g of CuCl₂

2.- Balanced chemical reaction

                  3CuCl₂  +  2 Al   ⇒    2AlCl₃  +  3Cu

3.- Calculate proportions

Theoretical proportion = (3 x 134.5)/(2 x 27) = 403.5/54 = 7.47

Experimental proportion = 167.5 g / 25 g = 6.7

4.- Conclusion

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CuCl₂.