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.
