Respuesta :
Answer: The mass percent of nitrogen gas in the compound is 13.3 %
Explanation:
Assuming the chemical equation of the compound forming product gases is:
[tex]\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)[/tex]
Now, the product gases are treated with KOH to remove carbon dioxide.
We are given:
[tex]p_{Total}=726torr\\P_{water}=23.8torr\\[/tex]
So, pressure of nitrogen gas will be = [tex]p_{Total}-p_{water}=726-23.8=702.2torr[/tex]
To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of nitrogen gas = 702.2 torr = 0.924 atm (Conversion factor: 1 atm = 760 torr)
V = Volume of nitrogen gas = 31.8 mL = 0.0318 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of nitrogen gas = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of nitrogen gas = ?
Putting values in above equation, we get:
[tex]0.924atm\times 0.0318L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\n_{mix}=\frac{0.924\times 0.0318}{0.0821\times 298}=0.0012mol[/tex]
- To calculate the mass of nitrogen gas, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of nitrogen gas = 28 g/mol
Moles of nitrogen gas = 0.0012 moles
Putting values in above equation, we get:
[tex]0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol\times 28g/mol)=0.0336g[/tex]
- To calculate the mass percent of nitrogen gas in compound, we use the equation:
[tex]\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}\times 100[/tex]
Mass of compound = 0.253 g
Mass of nitrogen gas = 0.0336 g
Putting values in above equation, we get:
[tex]\text{Mass percent of nitrogen gas}=\frac{0.0336g}{0.253g}\times 100=13.3\%[/tex]
Hence, the mass percent of nitrogen gas in the compound is 13.3 %
