Answer to Question #115914 in Organic Chemistry for Cici

Question

Aluminum bromide is produced by the reaction of aluminum metal with bromine liquid, according to the following balanced equation:

2 Al(s) + 3 Br2(l) → 2 AlBr3(aq)

If 10.0 grams of Al is added to 10.0 g of Br2, assuming the reaction has yielded 100% yield, calculate the mass (in grams) of AlBr3 produced in this reaction

Answer 1

Solution.

$2Al + 3Br2 = 2AlBr3$

$n(Al) = \frac{10}{26.98} = 0.37 \ mol$

$n(Br2) = \frac{10}{159.80}=0.06 \ mol$

In this reaction, Br2 is limiting reagent

$n(AlBr3) = \frac{0.06 \times 2}{3} = 0.04 \ mol$

Because yield of the reaction is 100 %, that:

$m(AlBr3) = 0.04 \times 266.68 = 10.67 \ g$

Answer:

m(AlBr3) = 10.67 g