Answer :
To determine the volume of [tex]\( \text{O}_2 \)[/tex] required to synthesize 15.0 mol of NO, we need to use the ideal gas law and stoichiometry of the chemical reaction involved. The chemical reaction involving NO is:
[tex]\[ 2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \][/tex]
From this reaction, we see that 2 moles of NO react with 1 mole of [tex]\( \text{O}_2 \)[/tex]. Therefore, we can calculate the moles of [tex]\( \text{O}_2 \)[/tex] required to synthesize 15.0 mol of NO.
1. Calculate the moles of [tex]\( \text{O}_2 \)[/tex] required:
[tex]\[ \text{Moles of } \text{O}_2 = \frac{15.0 \text{ mol NO}}{2} = 7.50 \text{ mol O}_2 \][/tex]
2. Convert the temperature from Celsius to Kelvin:
[tex]\[ T = 21 \, ^\circ \text{C} + 273.15 = 294.15 \, \text{K} \][/tex]
3. Convert the pressure from mmHg to atm:
[tex]\[ P = \frac{988 \, \text{mmHg}}{760 \, \text{mmHg per atm}} = 1.300 \, \text{atm} \][/tex]
4. Use the ideal gas law to calculate the volume of [tex]\( \text{O}_2 \)[/tex]:
The ideal gas law is given by:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure (1.300 atm)
- [tex]\( V \)[/tex] is the volume (unknown, this is what we are solving for)
- [tex]\( n \)[/tex] is the number of moles (7.50 mol)
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L·atm/(K·mol))
- [tex]\( T \)[/tex] is the temperature in Kelvin (294.15 K)
Rearranging the ideal gas law to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
[tex]\[ V = \frac{7.50 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 294.15 \, \text{K}}{1.300 \, \text{atm}} \][/tex]
5. Perform the calculation:
[tex]\[ V = \frac{7.50 \times 0.0821 \times 294.15}{1.300} \][/tex]
[tex]\[ V = \frac{181.06}{1.300} = 139.278 \, \text{L} \][/tex]
Rounded to three significant figures:
[tex]\[ V = 139 \, \text{L} \][/tex]
Answer:
The volume of [tex]\( \text{O}_2 \)[/tex] required to synthesize 15.0 mol of NO at 988 mmHg and 21 °C is [tex]\( 139 \, \text{L} \)[/tex].
[tex]\[ 2 \text{NO} + \text{O}_2 \rightarrow 2 \text{NO}_2 \][/tex]
From this reaction, we see that 2 moles of NO react with 1 mole of [tex]\( \text{O}_2 \)[/tex]. Therefore, we can calculate the moles of [tex]\( \text{O}_2 \)[/tex] required to synthesize 15.0 mol of NO.
1. Calculate the moles of [tex]\( \text{O}_2 \)[/tex] required:
[tex]\[ \text{Moles of } \text{O}_2 = \frac{15.0 \text{ mol NO}}{2} = 7.50 \text{ mol O}_2 \][/tex]
2. Convert the temperature from Celsius to Kelvin:
[tex]\[ T = 21 \, ^\circ \text{C} + 273.15 = 294.15 \, \text{K} \][/tex]
3. Convert the pressure from mmHg to atm:
[tex]\[ P = \frac{988 \, \text{mmHg}}{760 \, \text{mmHg per atm}} = 1.300 \, \text{atm} \][/tex]
4. Use the ideal gas law to calculate the volume of [tex]\( \text{O}_2 \)[/tex]:
The ideal gas law is given by:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure (1.300 atm)
- [tex]\( V \)[/tex] is the volume (unknown, this is what we are solving for)
- [tex]\( n \)[/tex] is the number of moles (7.50 mol)
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L·atm/(K·mol))
- [tex]\( T \)[/tex] is the temperature in Kelvin (294.15 K)
Rearranging the ideal gas law to solve for volume [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
[tex]\[ V = \frac{7.50 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 294.15 \, \text{K}}{1.300 \, \text{atm}} \][/tex]
5. Perform the calculation:
[tex]\[ V = \frac{7.50 \times 0.0821 \times 294.15}{1.300} \][/tex]
[tex]\[ V = \frac{181.06}{1.300} = 139.278 \, \text{L} \][/tex]
Rounded to three significant figures:
[tex]\[ V = 139 \, \text{L} \][/tex]
Answer:
The volume of [tex]\( \text{O}_2 \)[/tex] required to synthesize 15.0 mol of NO at 988 mmHg and 21 °C is [tex]\( 139 \, \text{L} \)[/tex].
