Question

An excess of Al 3.0 mol of Br2 are reacted according to the equation: 2Al +3Br2 = 2AlBr3
How many moles or AlBr3 will be formes assuming 100% yield?

Answer 1

so aluminum is in excess and you have 3.0 moles of Br2?

Answer 2

I believe so, I don't really understand it at all!

Answer 3

so, you can only calculate the number of moles you get in your products from the limiting reagent.

YOU are told that aluminum is in excess, so this means that Br2 will run out first.

so we use Br2 to find out how many moles of products we have

make sense?

Answer 4

Yes.

Answer 5

now, look at the equation is it balanced?

Answer 6

Or in other words are the number of atoms of each element the same on both the reactant and product side?

Answer 7

no

Answer 8

2Al +3Br2 = 2AlBr3

here's our equation

we have 2 moles of Aluminum on one side also 2 atoms of Aluminum

on the product side we have

2 moles moles of aluminum and 6 molecules of Br

Answer 9

FYI this is balanced

Answer 10

Yeah... how do i figure out how many moles of AlBr3 i will end up with?

Answer 11

so here's the key. To do that you must use the limiting reagent

Answer 12

we know that we have 3.0 mol of Br 2 right?

Answer 13

now we multiply this by the molar ratio of Br2 to AlBr3 like this.

This is the number of moles of AlBr3 if we had 100% yield

\[3.0 mol Br_{2}*(\frac{ 2AlBr_{3} }{ 3Br_{2} }) = 2 mol, of AlBr_{3}\]