Question

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

2 Al + 6HCl → 2 AlCl3 + 3 H2
Select one:
a. Al is the limiting reactant, 9.0 mol H2 can be formed
b. HCl is the limiting reactant, 6.5 mol H2 can be formed
c. Al is the limiting reactant, 6.0 mol H2 can be formed
d. HCl is the limiting reactant, 4.3 mol H2 can be formed

Answer 1

So we know from the balanced equation that in order for one entire reaction to occur, we need 2 moles of Al and 6 moles of HCl.

We are given 6.0 moles of Al, so let's divide by the amount we need for one reaction:

`(6)/(2) =3 reactions`

And we are given 13.0 moles of HCl. So let's divide this amount by how much we need for one reaction:

`(13)/(6)=2.1667 reactions`

So we know that this amount of HCl will give us less reactions, and therefore it is our limiting reactant. Now we must find the number of moles of
`H_(2)` that are produced with 13 moles of HCl.

We know that the molar ratio of HCl to
`H_(2)` is 6:3 in the balanced equation, so we can set up a proportion in order to find this new amount of
`H_(2)` that is produced:

`\frac{6_(HCl)}{3_{H_(2)}} =\frac{13_(HCl)}{x_{H_(2)}}`

And now we solve for x:

`6x=39`

`x=6.5mol_{H_(2)}`

So now we know that 6.5 moles of
`H_(2)` will be produced.

Therefore, our answer is B) HCl is the limiting reactant, 6.5 moles of
`H_(2)` can be formed.