Problem
You are given the following reaction:
2 H2(g) + O2(g) → 2 H2O(l)
Calculate:
a. the stoichiometric ratio of moles H2 to moles O2
b. the actual moles H2 to moles O2 when 1.50 mol H2 is mixed with 1.00 mol O2
c. the limiting reactant (H2 or O2) for the mixture in (b)
d. the theoretical yield, in moles, of H2O for the mixture in (b)
Solution
a. The stoichiometric ratio is given by using the coefficients of the balanced equation. The coefficients are the numbers listed before each formula. This equation is already balanced, so refer to the tutorial on balancing equations if you need further help:
2 mol H2 / mol O2
b. The actual ratio refers to the number of moles actually provided for the reaction. This may or may not be the same as the stoichiometric ratio. In this case, it is different:
1.50 mol H2 / 1.00 mol O2 = 1.50 mol H2 / mol O2
c. Note that the actual ratio of smaller than the required or stoichiometric ratio, which means there is insufficient H2 to react with all of the O2 that has been provided. The 'insufficient' component (H2) is the limiting reactant. Another way to put it is to say that O2 is in excess. When the reaction has proceeded to completion, all of the H2 will have been consumed, leaving some O2 and the product, H2O.
d. Theoretical yield is based on the calculation using the amount of limiting reactant, 1.50 mol H2. Given that 2 mol H2 forms 2 mol H2O, we get:
theoretical yield H2O = 1.50 mol H2 x 2 mol H2O / 2 mol H2
theoretical yield H2O = 1.50 mol H2O
Note that the only requirement for performing this calculation is knowing the amount of the limiting reactant and the ratio of the amount of limiting reactant to the amount of product.
Answer
a. 2 mol H2 / mol O2
b. 1.50 mol H2 / mol O2
c. H2
d. 1.50 mol H2O
