Question

What is the limiting reactant if 0.5 g al is reacted with 3.5 g cucl2? take into account cucl2 is a dihydrate when calculating the molecular weight?

Answer 1

Answer is: CuCl₂·2H₂O is limiting reactant.
Chemical reaction: 3CuCl₂·2H₂O + 2Al → 3Cu + 2AlCl₃ + 6H₂O.
m(Al) = 0,5 g.
m(CuCl₂·2H₂O) = 3,5 g.
n(CuCl₂·2H₂O) = m(CuCl₂·2H₂O) ÷ M(CuCl₂·2H₂O).
n(CuCl₂·2H₂O) = 3,5 g ÷ 170,5 g/mol.
n(CuCl₂·2H₂O) = 0,0205 mol.
n(Al) = 0,5 g ÷ 27 g/mol.
n(Al) = 0,0185 mol.
From chemical reaction: n(Al) : n(CuCl₂·2H₂O) = 2 : 3.
0,0185 mol : n(CuCl₂·2H₂O) = 2 : 3.
n(CuCl₂·2H₂O) = 0,02755 mol; there is no enough n(CuCl₂·2H₂O).

Answer 2

The limiting reactant for the reaction is CuCl₂•2H₂O

Balanced equation

3CuCl₂•2H₂O + 2Al → 3Cu + 2AlCl₃ + 6H₂O

Molar mass of CuCl₂•2H₂O = 63.5 + (35.5×2) + 2[(2×1) + 16] = 170.5 g/mol

Mass of CuCl₂•2H₂O from the balanced equation = 3 × 170.5 = 511.5 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

SUMMARY

From the balanced equation above,

511.5 g of CuCl₂•2H₂O reacted with 54 g of Al

How to determine the limiting reactant

From the balanced equation above,

511.5 g of CuCl₂•2H₂O reacted with 54 g of Al.

Therefore,

3.5 g of CuCl₂•2H₂O will react with = (3.5 × 54) / 511.5 = 0.4 g of Al

From the calculation made above, we can see that only 0.4 g of Al out of 0.5 g given is needed to react completely with 3.5 g of CuCl₂•2H₂O.

Therefore, CuCl₂•2H₂O is the limiting reactant

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