Question

A mixture of nitrogen and hydrogen gases, at a total pressure of 663 mm Hg, contains 3.46 grams of nitrogen and 0.156 grams of hydrogen. What is the partial pressure of each gas in the mixture? PN2 = mm Hg PH2 = mm Hg

Answer 1

Answer: The partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg

Step-by-step explanation:

To calculate the number of moles, we use the equation:

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

• For nitrogen gas:

Given mass of nitrogen gas = 3.46 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

`\text{Moles of nitrogen gas}=(3.46g)/(28g/mol)=0.123mol`

• For hydrogen gas:

Given mass of hydrogen gas = 0.156 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

`\text{Moles of hydrogen gas}=(0.156g)/(2g/mol)=0.078mol`

Mole fraction of a gas is calculated by using the formula:

`\chi_(A)=(n_(A))/(n_(A)+n_(B))` ......(1)

• For nitrogen gas:

Putting values in equation 1, we get:

`\chi_{\text{nitrogen gas}}=(0.123)/(0.123+0.078)=0.612`

• For hydrogen gas:

Putting values in equation 1, we get:

`\chi_{\text{hydrogen gas}}=(0.078)/(0.123+0.078)=0.388`

The partial pressure of a gas is given by Raoult's law, which is:

`p_A=p_T* \chi_A` ......(2)

where,

`p_A` = partial pressure of substance A

`p_T` = total pressure = 663 mmHg

`\chi_A` = mole fraction of substance A

• For nitrogen gas:

`p_{\text{Nitrogen gas}}=663mmHg* 0.612\\\\p_{\text{Nitrogen gas}}=405.76mmHg`

• For hydrogen gas:

`p_{\text{Hydrogen gas}}=663mmHg* 0.388\\\\p_{\text{Hydrogen gas}}=257.24mmHg`

Hence, the partial pressure of nitrogen gas is 405.76 mmHg and that of hydrogen gas is 257.24 mmHg