Based on the Lewis/electron dot representation
1 answer:
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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.
<u>Answer:</u> The mole fraction of gas B (hydrogen gas) is 0.557
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For nitrogen gas:</u>
Given mass of nitrogen gas = 10.25 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
- <u>For hydrogen gas:</u>
Given mass of hydrogen gas = 2.05 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
- <u>For ammonia gas:</u>
Given mass of ammonia gas = 7.63 g
Molar mass of ammonia gas = 17 g/mol
Putting values in equation 1, we get:
Mole fraction of a substance is given by:
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:
Hence, the mole fraction of gas B (hydrogen gas) is 0.557