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:
[tex] \frac{6}{2} =3 reactions[/tex]
And we are given 13.0 moles of HCl. So let's divide this amount by how much we need for one reaction:
[tex] \frac{13}{6}=2.1667 reactions [/tex]
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 [tex] H_{2} [/tex] that are produced with 13 moles of HCl.
We know that the molar ratio of HCl to [tex] H_{2} [/tex] is 6:3 in the balanced equation, so we can set up a proportion in order to find this new amount of [tex] H_{2} [/tex] that is produced:
[tex] \frac{6_{HCl}}{3_{H_{2}}} =\frac{13_{HCl}}{x_{H_{2}}} [/tex]
And now we solve for x:
[tex] 6x=39 [/tex]
[tex] x=6.5mol_{H_{2}} [/tex]
So now we know that 6.5 moles of [tex] H_{2} [/tex] will be produced.
Therefore, our answer is B) HCl is the limiting reactant, 6.5 moles of [tex] H_{2} [/tex] can be formed.
