Question

The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the KOH solution, the gas contains N2 and is saturated with water vapor. In a given experiment a 0.253-g sample of a compound produced 31.8 mL N2 saturated with water vapor at 258C and 726 torr. What is the mass percent of nitrogen in the compound

Answer 1

Answer: The mass percent of nitrogen gas in the compound is 13.3 %

Step-by-step explanation:

Assuming the chemical equation of the compound forming product gases is:

`\text{Compound}\xrightarrow[CuO(s)]{Hot}N_2(g)+CO_2(g)+H_2O(g)`

Now, the product gases are treated with KOH to remove carbon dioxide.

We are given:

`p_(Total)=726torr\\P_(water)=23.8torr\\`

So, pressure of nitrogen gas will be =
`p_(Total)-p_(water)=726-23.8=702.2torr`

To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:

`PV=nRT`

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 =
`25^oC=[25+273]K=298K`

R = Gas constant =
`0.0821\text{ L. atm }mol^(-1)K^(-1)`

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

`0.924atm* 0.0318L=n* 0.0821\text{ L atm }mol^(-1)K^(-1)* 298K\\n_(mix)=(0.924* 0.0318)/(0.0821* 298)=0.0012mol`

• To calculate the mass of nitrogen gas, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.0012 moles

Putting values in above equation, we get:

`0.0012mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.0012mol* 28g/mol)=0.0336g`

• To calculate the mass percent of nitrogen gas in compound, we use the equation:

`\text{Mass percent of nitrogen gas}=\frac{\text{Mass of nitrogen gas}}{\text{Mass of compound}}* 100`

Mass of compound = 0.253 g

Mass of nitrogen gas = 0.0336 g

Putting values in above equation, we get:

`\text{Mass percent of nitrogen gas}=(0.0336g)/(0.253g)* 100=13.3\%`

Hence, the mass percent of nitrogen gas in the compound is 13.3 %