Chemistry / Unit 8: Stoichiometry
8.1 Mole Ratios
Notes
Stoichiometry uses a balanced equation to relate amounts of reactants and products. The coefficients give the mole ratio.
- In $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$, the ratios are 1 mol $\text{N}_2$ : 3 mol $\text{H}_2$ : 2 mol $\text{NH}_3$.
- Moles of B $=$ moles of A $\displaystyle \times \frac{\text{coefficient of B}}{\text{coefficient of A}}$.
- Example: 6.0 mol $\displaystyle \text{H}_2 \times \frac{2 \text{ mol NH}_3}{3 \text{ mol H}_2} = 4.0$ mol $\text{NH}_3$.
- The equation must be balanced first. Coefficients relate moles, not grams.
Practice
In $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, how many moles of $\text{O}_2$ are needed to react with 5.0 mol of $\text{H}_2$?
$\displaystyle 5.0 \text{ mol H}_2 \times \frac{1 \text{ mol O}_2}{2 \text{ mol H}_2} = 2.5$ mol.
In $2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$, how many moles of $\text{O}_2$ are needed to burn 4.0 mol of $\text{C}_2\text{H}_6$?
$\displaystyle 4.0 \times \frac{7}{2} = 14$ mol $\text{O}_2$.
In $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$, how many moles of water form when 3.0 mol of $\text{CO}_2$ form?
The ratio of $\text{H}_2\text{O}$ to $\text{CO}_2$ is $2 : 1$, so $3.0 \times 2 = 6.0$ mol.
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