Question

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?

Answer 1

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

Answer 2

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