Respuesta :
Answer:
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.
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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