Question

How many moles of Al are necessary to form 23.6 g of AlBr3 ---> 2 AlBr3

Answer 1

Answer:

$n_{Al}=0.0885molAl$

Explanation:

Hello,

In this case, the undergoing chemical reaction should be:

$2Al+3Br_2\rightarrow 2AlBr_3$

In such a way, since aluminum bromide molar mass is 266.7 g/mol and it is in a 2:2 molar ratio with aluminum, we compute the necessary moles of aluminum as shown below:

$n_{Al}=23.6gAlBr_3*\frac{1molAlBr_3}{266.7gAlBr_3} *\frac{2molAl}{2molAlBr_3} \\\\n_{Al}=0.0885molAl$

Best regards.

Answer 2

Answer:

0.088 mole

Explanation:

0.088 mole of Al would be necessary.

From the balanced equation of reaction:

$2 Al (s) + 3 Br_2 (l) --> 2 AlBr_3$

The mole ratio of Al to $AlBr_3$ is 1:1, meaning that 1 mole of the former is required to produce 1 mole of the latter.

Recall that: mole = mass/molar mass

Molar mass of $AlBr_3$ = 266.7 g/mol

23.6 g of $AlBr_3$ = 23.6/266.7 = 0.088 mole

Since the mole ratio of the two compounds according to the balanced equation of reaction is 1:1, it thus means that 0.088 moles of Al will also be needed for the reaction.

Hence, 0.088 mole of Al are necessary to produce 23.6 g of $AlBr_3$