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

Explanation:

Hello,

In this case, the undergoing chemical reaction should be:

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:

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0.088 mole

Explanation:

0.088 mole of Al would be necessary.

From the balanced equation of reaction:

The mole ratio of Al to 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 = 266.7 g/mol

23.6 g of = 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