Question

When 0.40 mol Al is mixed with 0.40 mol Br2, the following reaction occurs: 2Al(s) + 3Br2(l) → 2AlBr3(s) Identify the limiting reactant and the moles of leftover reactant, assuming the reaction goes to completion.

a. The limiting reactant is Br2, and 0.20 mol Al should remain unreacted.
b. The limiting reactant is Al, and 0.13 mol Br2 should remain unreacted.
c. The limiting reactant is Al, and 0.20 mol Br2 should remain unreacted.
d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.

Answer 1

Answer:

d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.

Explanation:

From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)

2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3

Hence 1 mole of Br combines with 2/3 moles of Al

or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving

0.4 - 0.267 = 0.133 moles of Al remaining unreacted