QuestionJuly 17, 2025

5) Determine the pressure (in mmHg) for the nitrogen gas sample that is collected when 53.0 g of NaN_(3) decomposes. The temperature of the gas is 25^circ C and the volume is 18.0 L. underline ( )NaN_(3)(s)arrow underline ( )Na(s)+underline ( )N_(2)(g)

5) Determine the pressure (in mmHg) for the nitrogen gas sample that is collected when 53.0 g of NaN_(3) decomposes. The temperature of the gas is 25^circ C and the volume is 18.0 L. underline ( )NaN_(3)(s)arrow underline ( )Na(s)+underline ( )N_(2)(g)
5) Determine the pressure (in mmHg) for the nitrogen gas sample that is collected when 53.0 g of NaN_(3) decomposes. The temperature of the gas is 25^circ C and the volume is 18.0 L. underline ( )NaN_(3)(s)arrow underline ( )Na(s)+underline ( )N_(2)(g)

Solution
4.0(269 votes)

Answer

1266.92 mmHg Explanation 1. Determine moles of NaN_3 Molar mass of NaN_3 is 65.01 g/mol. Moles = \frac{53.0 \text{ g}}{65.01 \text{ g/mol}} = 0.815 \text{ mol}. 2. Calculate moles of N_2 produced From the balanced equation, 2 moles of NaN_3 produce 3 moles of N_2. Moles of N_2 = 0.815 \times \frac{3}{2} = 1.2225 \text{ mol}. 3. Use Ideal Gas Law to find pressure Convert temperature to Kelvin: 25^{\circ}C + 273.15 = 298.15 \text{ K}. **Ideal Gas Law**: PV = nRT, solve for P: P = \frac{nRT}{V}. Use R = 0.0821 \text{ L atm/mol K}, n = 1.2225 \text{ mol}, T = 298.15 \text{ K}, V = 18.0 \text{ L}. P = \frac{1.2225 \times 0.0821 \times 298.15}{18.0} = 1.667 \text{ atm}. 4. Convert pressure to mmHg 1 \text{ atm} = 760 \text{ mmHg}. Pressure in mmHg = 1.667 \times 760 = 1266.92 \text{ mmHg}.

Explanation

1. Determine moles of $NaN_3$<br /> Molar mass of $NaN_3$ is 65.01 g/mol. Moles = $\frac{53.0 \text{ g}}{65.01 \text{ g/mol}} = 0.815 \text{ mol}$.<br /><br />2. Calculate moles of $N_2$ produced<br /> From the balanced equation, 2 moles of $NaN_3$ produce 3 moles of $N_2$. Moles of $N_2$ = $0.815 \times \frac{3}{2} = 1.2225 \text{ mol}$.<br /><br />3. Use Ideal Gas Law to find pressure<br /> Convert temperature to Kelvin: $25^{\circ}C + 273.15 = 298.15 \text{ K}$.<br /> **Ideal Gas Law**: $PV = nRT$, solve for $P$: $P = \frac{nRT}{V}$.<br /> Use $R = 0.0821 \text{ L atm/mol K}$, $n = 1.2225 \text{ mol}$, $T = 298.15 \text{ K}$, $V = 18.0 \text{ L}$.<br /> $P = \frac{1.2225 \times 0.0821 \times 298.15}{18.0} = 1.667 \text{ atm}$.<br /><br />4. Convert pressure to mmHg<br /> $1 \text{ atm} = 760 \text{ mmHg}$. Pressure in mmHg = $1.667 \times 760 = 1266.92 \text{ mmHg}$.
Click to rate:

Similar Questions