Respuesta :
Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.
Explanation:
To calculate the number of moles, we use the equation given by ideal gas equation:
PV = nRT
where,
P = Pressure of the gaseous mixture = 1.00 atm
V = Volume of the gaseous mixture = 24.62 L
n = number of moles of the gaseous mixture = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gaseous mixture = 300 K
Putting values in above equation, we get:
[tex]1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol[/tex]
We are given:
Total mass of the mixture = 13.22 grams
Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For nitrogen gas:
Molar mass of nitrogen gas = 28 g/mol
[tex]\text{Moles of nitrogen gas}=\frac{x}{28}mol[/tex]
For hydrogen gas:
Molar mass of hydrogen gas = 2 g/mol
[tex]\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol[/tex]
Equating the moles of the individual gases to the moles of mixture:
[tex]0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g[/tex]
To calculate the mass percentage of substance in mixture we use the equation:
[tex]\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100[/tex]
Mass of the mixture = 13.22 g
- For nitrogen gas:
Mass of nitrogen gas = x = 12.084 g
Putting values in above equation, we get:
[tex]\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%[/tex]
- For hydrogen gas:
Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g
Putting values in above equation, we get:
[tex]\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%[/tex]
Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.
