Question

a mixture of gases contains 2.14 g of $n_2$, 5.85 g of $h_2$, and 4.18 g of $nh_3$. if the total pressure of the mixture is 4.58 atm, what is the partial pressure of each component?

Explanation:

Step1: Calculate moles of each gas

Molar mass of $\text{N}_2$ = 28.02 g/mol, $\text{H}_2$ = 2.016 g/mol, $\text{NH}_3$ = 17.03 g/mol
$n_{\text{N}_2} = \frac{2.14\ \text{g}}{28.02\ \text{g/mol}} \approx 0.0764\ \text{mol}$
$n_{\text{H}_2} = \frac{5.85\ \text{g}}{2.016\ \text{g/mol}} \approx 2.902\ \text{mol}$
$n_{\text{NH}_3} = \frac{4.18\ \text{g}}{17.03\ \text{g/mol}} \approx 0.2455\ \text{mol}$

Step2: Find total moles of gas

$n_{\text{total}} = n_{\text{N}_2} + n_{\text{H}_2} + n_{\text{NH}_3}$
$n_{\text{total}} = 0.0764 + 2.902 + 0.2455 \approx 3.2239\ \text{mol}$

Step3: Calculate mole fractions

$X_{\text{N}_2} = \frac{n_{\text{N}_2}}{n_{\text{total}}} = \frac{0.0764}{3.2239} \approx 0.0237$
$X_{\text{H}_2} = \frac{n_{\text{H}_2}}{n_{\text{total}}} = \frac{2.902}{3.2239} \approx 0.9002$
$X_{\text{NH}_3} = \frac{n_{\text{NH}_3}}{n_{\text{total}}} = \frac{0.2455}{3.2239} \approx 0.0762$

Step4: Find partial pressures ($P_i = X_i \times P_{\text{total}}$)

$P_{\text{N}_2} = 0.0237 \times 4.58\ \text{atm} \approx 0.109\ \text{atm}$
$P_{\text{H}_2} = 0.9002 \times 4.58\ \text{atm} \approx 4.12\ \text{atm}$
$P_{\text{NH}_3} = 0.0762 \times 4.58\ \text{atm} \approx 0.349\ \text{atm}$

Answer:

Partial pressure of $\text{N}_2$: $\approx 0.109$ atm
Partial pressure of $\text{H}_2$: $\approx 4.12$ atm
Partial pressure of $\text{NH}_3$: $\approx 0.349$ atm