Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
6.5 mol of O2 will be left.
Balanced equation — do this first, always
2 H2(g) + O2(g) → 2 H2O(g)
The 2:1 ratio is the entire problem. Skip the balancing and you will subtract 5.0 from 9.0 and answer 4.0, which is the most common wrong answer here.
Step 1 — how much O2 does 5.0 mol of H2 consume?
The equation says every 2 mol H2 uses 1 mol O2, so:
5.0 mol H2 × (1 mol O2 / 2 mol H2) = 2.5 mol O2 consumed
Step 2 — subtract from what you started with
9.0 mol − 2.5 mol = 6.5 mol O2 remaining
Reading the question correctly
"Suppose as much as possible of the H2 reacts" is the problem telling you outright that H2 is limiting — all 5.0 mol of it is consumed, and 0 mol H2 is left. So "how much will be left" has to mean the excess reagent, the O2.
You can confirm H2 really is limiting: divide each amount by its coefficient. H2: 5.0/2 = 2.5. O2: 9.0/1 = 9.0. The smaller quotient is the limiting reactant, so H2 runs out first, comfortably. That divide-by-the-coefficient check is the reliable way to identify a limiting reactant — comparing raw mole counts only works when the coefficients happen to be equal.
Full accounting when the reaction stops:
H2: 0 mol | O2: 6.5 mol | H2O: 5.0 mol
(Water comes out 1:1 with H2, since both carry a coefficient of 2.)
Quick sanity check on the answer: you started with 14.0 mol of gas total and ended with 11.5 mol. That drop is expected — the reaction consumes three molecules and produces two, so total moles must fall. If your numbers had gone up, something was wrong.
