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An excess of al and 6.0 mol of br2 are reacted according to the equation 2al + 3br2 --> 2albr3 how many moles of albr3 will be formed assuming 100% yield?

User Sum NL
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2 Answers

5 votes

Answer : The number of moles of
AlBr_3 formed will be, 4 moles

Explanation : Given,

Moles of
Br_2 = 6.0 mole

The given balanced chemical reaction is,


2Al+3Br_2\rightarrow 2AlBr_3

From the balanced chemical reaction, we conclude that

As, 3 moles of
Br_2 react to give 2 moles of
AlBr_3

So, 6 moles of
Br_2 react to give
(2)/(3)* 6=4 moles of
AlBr_3

Therefore, the number of moles of
AlBr_3 formed will be, 4 moles

User Ethan Wu
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6.6k points
4 votes
The reactant has excess aluminum so bromine would be the limiting reagent. Looking at the coefficient of the equation, every 3 mol Br2 will produce 2 mol of AlBr3.
So, the amount of AlBr3 produced by 6 mol of Br2 would be:

Br2/AlBr3= 3/2
6mol/AlBr3= 3/2
AlBr3= 6mol * 2/3= 4 mol
User Umberto
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7.2k points