For the reaction 2H2 + O2 → 2H2O, you mix 4 mol of H2 with 1 mol of O2. What limits the water you can make?

The hydrogen, there is more of it, so it dominates

Nothing, all 5 mol of gas become water

The reaction stops at 1 mol of water

Physics I / Stoichiometric relationships

Practice question · Multiple choice

For the reaction $2H_{2} + O_{2} \to 2H_{2}O$ , you mix 4 mol of $H_2$ with 1 mol of $O_2$ . What limits the water you can make?

Hints

The recipe demands a 2:1 ratio of $H_2$ to $O_2$ . Compare what each ingredient could produce alone.
1 mol $O_2$ can partner only 2 mol $H_2$ , the rest of the hydrogen has no oxygen to react with.

Show the answer

C. The oxygen, leaving 2 mol of H₂ unused

Why

Stoichiometry is recipe arithmetic: 1 mol O₂ supports exactly 2 mol H₂ → 2 mol H₂O, stranding the excess hydrogen. The scarcer-in-ratio ingredient (the limiting reagent) fixes the yield, like running out of cheese with bread to spare.

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For the reaction 2H2+O2→2H2O2H_{2} + O_{2} \to 2H_{2}O2H2​+O2​→2H2​O, you mix 4 mol of H2H_2H2​ with 1 mol of O2O_2O2​. What limits the water you can make?

Practise Stoichiometric relationships

More questions on Stoichiometric relationships

For the reaction $2H_{2} + O_{2} \to 2H_{2}O$ , you mix 4 mol of $H_2$ with 1 mol of $O_2$ . What limits the water you can make?