In the reaction 2B + 3H2 --> B2H6, 9.422 moles B reacts with 14.102 moles of H2, what is the limiting reactant?

Start by taking a look at the balanced chemical equation for this reaction

#color(red)(2)"B"_text((s]) + color(blue)(3)"H"_text(2(g]) -> "B"_2"H"_text(6(g])#

Notice that you have a #color(red)(2): color(blue)(3)# mole ratio between boron and hydrogen gas. This means that the two reactants will always take part in the reaction in this proportion.

That is, regardless of how many moles of boron you have, you will always need at least #3/2# times more moles of hydrogen gas. LIkewise, regardless of how many moles of hydrogen gas you have, you will always need at least #2/3# time less moles of boron.

So, how many moles of hydrogen gas would you need in order to make sure that all the moles of boron react?

#9.422color(red)(cancel(color(black)("moles B"))) * (color(blue)(3)color(white)(x)"moles H"_2)/(color(red)(2)color(red)(cancel(color(black)("moles B")))) = "14.133 moles H"_2#

Notice that this almost equal to the number of moles of hydrogen gas given to you.

Since the number of moles of hydrogen gas given to you is smaller than the number of moles of hydrogen gas you would need to make sure that all the boron reacts, it follows that hydrogen gas will be the limiting reagent.

This is equivalent to saying the boron will be in excess. More specifically, the reaction will consume all the moles of hydrogen gas and

#14.102color(red)(cancel(color(black)("moles H"_2))) * (color(red)(2)color(white)(x)"moles B")/(color(blue)(3)color(red)(cancel(color(black)("moles H"_2)))) = "9.4013 moles B"#

The number of moles of boron that will remain in excess is

#9.422-9.4013 = "0.021 moles B" -># in excess