Upper secondary · 15 min

Stoichiometry: limiting reagent and yield

Vary the starting amounts of $\ce{H2}$ and $\ce{O2}$ and see which one limits the amount of water formed.

Goal

Use the balanced equation $\ce{2H2 + O2 -> 2H2O}$ to identify the limiting reagent and predict the theoretical yield in each mixture.

Apparatus and reagents

Hydrogen and oxygen gas cylinders (simulated), reaction vessel, balanced equation $\ce{2H2 + O2 -> 2H2O}$ , calculator.

Procedure

Select “Exact ratio 2 : 1”: read the heights of the $\ce{H2}$ , $\ce{O2}$ and $\ce{H2O}$ vessels and check they match the coefficients.
Switch to “Hydrogen in excess”: identify which reactant empties completely and how much of the other remains.
Switch to “Oxygen in excess” and repeat the analysis.
For each mode compute the theoretical moles of $\ce{H2O}$ from the limiting reagent, then compare with the product vessel.

What to observe

At the exact 2 : 1 ratio both reactant vessels empty completely and the product reaches its maximum.
In excess modes the limiting vessel empties first; the leftover reactant cannot form more product.

Explanation

The coefficients of $\ce{2H2 + O2 -> 2H2O}$ fix the mole ratio 2 : 1 : 2. The reactant present in the smaller stoichiometric amount — the limiting reagent — sets the theoretical yield; the other remains as excess. Actual yield is often lower and reported as a percentage of theory.