Chemistry, 22.12.2020 02:40 jessecabrown1

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?

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Answer from: lilpeepxliltracy

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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Answer from: stormhorn491

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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Aluminum bromide can be prepared by the reaction of aluminum metal with bromine gas shown by the equ...