Practice #1 – Limiting Reactants in Moles

Question

    How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?

2H2+O2--> 2 H2O

Answer 1

One way to find the limiting reactant is to divide the moles of each reactant by the coefficient of that reactant in the balanced equation. In this problem, we will divide moles of H2 by 2 and moles of O2 by 1.

H₂: 4.3 mols / 2 = 2.15

O₂: 5.6 mols / 1 = 5.6

2.15 is less than 5.6 so H₂ is limiting.

Now we use the moles (not moles divided by 2) of H₂ to find moles H₂O formed.

4.3 mols H₂ x 2 mol H₂O / 2 mol H₂ = 4.3 moles H₂O

To find moles O₂ left over, we use dimensional analysis:

4.3 mols H₂ x 1 mol O₂ / 2 mols H₂ = 2.15 mols O₂ used up

moles left over = 5.6 mols - 2.15 mols = 3.45 mols O₂ left over = 3.5 mols O₂ left over (2 sig. figs.)

Answer 2

2 moles of H2 reacts with 1 of O2

So 1 mole reacts with 0.5

So 4.3 moles of H2 reacts with 4.3 x 0.5 = 2.15 moles of O2

So the H2 is limiting as there is insufficient to react with all of the O2

5.6 - 2.15 = 3.45 moles of O2 will be left

Number of moles of water = number of moles of H2 reacting = 4.3