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?

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:

$\frac{6}{2} =3 reactions$

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

$\frac{13}{6}=2.1667 reactions$

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.