Chemistry / Unit 8: Stoichiometry
8.3 Limiting Reactant
Notes
The limiting reactant runs out first and decides how much product forms. The other is the excess reactant.
- Method: convert each reactant to moles, then find how much product each could make. The one that makes less product is limiting.
- Or: find how much of B is needed to use up all of A, and compare with what you have.
- Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with 3.0 mol $\text{H}_2$ and 2.0 mol $\text{O}_2$. 3.0 mol $\text{H}_2$ needs only 1.5 mol $\text{O}_2$, so $\text{H}_2$ is limiting. It makes 3.0 mol $\text{H}_2\text{O}$, and 0.5 mol $\text{O}_2$ is left over.
- The limiting reactant is not always the one with the smaller mass or fewer moles.
Practice
In $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$, 2.0 mol of $\text{N}_2$ reacts with 3.0 mol of $\text{H}_2$. How many moles of $\text{NH}_3$ can form?
2.0 mol $\text{N}_2$ would need 6.0 mol $\text{H}_2$, so $\text{H}_2$ is limiting. $\displaystyle 3.0 \times \frac{2}{3} = 2.0$ mol $\text{NH}_3$.
In $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$, a student uses 13.08 g of Zn (65.38 g/mol) and 7.30 g of HCl (36.46 g/mol). Which is limiting, and how many moles of $\text{H}_2$ form?
Zn: 0.200 mol; HCl: 0.200 mol. All the Zn would need 0.400 mol HCl, so HCl is limiting. $\displaystyle 0.200 \text{ mol HCl} \times \frac{1}{2} = 0.100$ mol $\text{H}_2$.
In $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, 4.0 mol of $\text{H}_2$ reacts with 4.0 mol of $\text{O}_2$. How many moles of the excess reactant remain?
4.0 mol $\text{H}_2$ uses 2.0 mol $\text{O}_2$, leaving $4.0 - 2.0 = 2.0$ mol $\text{O}_2$.
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