Respuesta :
The question is incomplete, here is the complete question:
Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?
A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.
Answer: The mole fraction of gas B (hydrogen gas) is 0.557
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For nitrogen gas:
Given mass of nitrogen gas = 10.25 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol[/tex]
- For hydrogen gas:
Given mass of hydrogen gas = 2.05 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol[/tex]
- For ammonia gas:
Given mass of ammonia gas = 7.63 g
Molar mass of ammonia gas = 17 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B+n_C}[/tex]
Moles of gas B (hydrogen gas) = 1.025 moles
Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles
Putting values in above equation, we get:
[tex]\chi_{(H_2)}=\frac{1.025}{1.84}=0.557[/tex]
Hence, the mole fraction of gas B (hydrogen gas) is 0.557
