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

Examine the balanced chemical equation for this reaction first.

In order to ensure that every mole of boron is reacted, how many moles of hydrogen gas would you need?

You'll see that this is nearly equal to the quantity of hydrogen gas you were given.

It follows that hydrogen gas will be the limiting reagent since the amount of hydrogen gas you are given is less than the amount of hydrogen gas you would need to ensure that all of the boron reacts.

This means there will be an excess of boron; more precisely, the reaction will use up all the moles of hydrogen gas and

The quantity of boron that will continue to be in excess is

To determine the limiting reactant, first, calculate the moles of B2H6 that would be produced if all reactants were fully consumed. Then, compare the moles of B2H6 produced from each reactant. The reactant that produces the lesser amount of B2H6 is the limiting reactant.

Using the stoichiometry of the reaction:

2 moles of B yields 1 mole of B2H6 3 moles of H2 yields 1 mole of B2H6

Calculate the moles of B2H6 produced from 9.422 moles of B and 14.102 moles of H2:

For B: (9.422 \text{ moles} \times \frac{1 \text{ mole B2H6}}{2 \text{ moles B}} = 4.711 \text{ moles B2H6})

For H2: (14.102 \text{ moles} \times \frac{1 \text{ mole B2H6}}{3 \text{ moles H2}} = 4.701 \text{ moles B2H6})

Since B produces fewer moles of B2H6 compared to H2, B is the limiting reactant.