Question

find the # of moles in each case:

1. 2.5 l of $o_2$
2. 67 l of $n_2$
3. 120 l of steam

find the volume of each

1. 3.4 moles of $o_2$
2. 1.25 moles of $n_2$

Explanation:

We assume standard temperature and pressure (STP) conditions, where 1 mole of any ideal gas occupies 22.4 L.

Step1: Moles of $\text{O}_2$ from volume

Divide volume by molar volume.
$\text{Moles of O}_2 = \frac{2.5\ \text{L}}{22.4\ \text{L/mol}}$

Step2: Moles of $\text{N}_2$ from volume

Divide volume by molar volume.
$\text{Moles of N}_2 = \frac{67\ \text{L}}{22.4\ \text{L/mol}}$

Step3: Moles of steam from volume

Divide volume by molar volume.
$\text{Moles of steam} = \frac{120\ \text{L}}{22.4\ \text{L/mol}}$

Step4: Volume of $\text{O}_2$ from moles

Multiply moles by molar volume.
$\text{Volume of O}_2 = 3.4\ \text{mol} \times 22.4\ \text{L/mol}$

Step5: Volume of $\text{N}_2$ from moles

Multiply moles by molar volume.
$\text{Volume of N}_2 = 1.25\ \text{mol} \times 22.4\ \text{L/mol}$

Answer:

1. Moles of $\text{O}_2$: $\frac{2.5}{22.4} \approx 0.11\ \text{mol}$
2. Moles of $\text{N}_2$: $\frac{67}{22.4} \approx 3.0\ \text{mol}$
3. Moles of steam: $\frac{120}{22.4} \approx 5.36\ \text{mol}$
4. Volume of $\text{O}_2$: $3.4 \times 22.4 = 76.16\ \text{L}$
5. Volume of $\text{N}_2$: $1.25 \times 22.4 = 28.0\ \text{L}$