Aluminum bromide can be prepared by the reaction of aluminum metal with bromine gas shown by the equation: 2 Al + 3 Br2 → 2 AIBr; Now suppose that 5.6 mol of aluminum reacts with 4.4 mol of bromine. How młuch aluminum bromide will be produced?

The amount of  aluminium bromide produced is 2.93 moles

The balance chemical equation can be represented as follows;

2Al + 3 Br₂ → 2AIBr₃

let's find the limiting reagent .

Therefore,

moles of Al = 5.6 / 2 = 2.8

moles of Br = 4.4 / 3 = 1.46666666667

Bromine is the limiting reagent. The limiting reagent determines the amount of product.

Therefore,

3 moles of bromine gives  2 moles of AIBr₃

4.4 moles will give ?

cross multiply

number of moles of  aluminium bromide = 4.4 × 2 / 3 = 8.8 / 3 = 2.93 moles

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2.93 moles of AIBr₃

Explanation:

The balanced equation for the reaction is given below:

2Al + 3Br₂ → 2AIBr₃

From the balanced equation above,

2 moles of Al reacted with 3 moles of Br₂ to produce 2 moles of AIBr₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

2 moles of Al reacted with 3 moles of Br₂.

Therefore, 5.6 moles of Al will react with = (5.6 × 3)/2 = 8.4 moles of Br₂

From the calculation made above, we can see clearly that it will take a higher amount (i.e 8.4 moles) of Br₂ than what was given (i.e 4.4 moles) to completely react with 5.6 moles of Al.

Therefore, Br₂ is the limiting reactant and Al is the excess reactant.

Finally, we shall determine the amount of AIBr₃ produced from the reaction.

In this case, we shall use the limiting reactant because it will produce the maximum amount of AIBr₃ since all of it is consumed in the reaction.

The limiting reactant is Br₂ and the amount of AIBr₃ produced can be obtained as follow:

From the balanced equation above,

3 moles of Br₂ reacted to produce 2 moles of AIBr₃.

Therefore, 4.4 moles of Br₂ will react to produce = (4.4 × 2)/3 = 2.93 moles of AIBr₃.

Thus, 2.93 moles of AIBr₃ is produced from the reaction.

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