If 25.0 L of HCl is reacted with 50.0 L of O2, what is the limiting reactant for 4 HCl (g) + O2 (g) → 2 Cl2 (g) + 2H2O (g)?

Sep 16, 2014

The limiting reactant is HCl.

You use the usual method for finding limiting reactants.

In your problem, the balanced equation is

4HCl(g) + O₂(g) → 2Cl₂(g) + 2H₂O(g)

According to Gay-Lussac's Law of Combining Volumes, gases react in fixed ratios by volume. We can re-write the equation as

4 L HCl(g) + 1 L O₂(g) → 2 L Cl₂(g) + 2 L H₂O(g)

So we can skip Step 2 and go directly to Step 3. We can use the gas volumes instead of moles to find the limiting reactant.

From HCl: 25.0 L HCl × ("2 LCl"_2)/"4 L HCl" = 12.5 mol Cl₂

From O₂: 50.0 L 0_2 × $\left({\text{2 L Cl"_2)/("1 L O}}_{2}\right)$ = 100. mol Cl₂

So the limiting reactant is HCl.