A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.70 mol of Gas A in the mixture, how many moles of Gas B are present?

Question

Chemistry

Camren Cowan

12 May, 15:59

A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.70 mol of Gas A in the mixture, how many moles of Gas B are present?

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Answers (1)

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Fern · Answer 1

moles B = 2.32 moles

Explanation:

In this case, we can assume that both gases are ideals, so we can use the expression for an ideal gas which is:

PV = nRT

From here, we can calculate the total moles (n) that are in the container, and then, by difference, we can calculate how much we have of gas B.

For this case, we will use R = 0.082 L atm / mol K. Solving for n:

n = PV/RT

n = 5 * 20 / 0.082 * 303

n = 4.02 moles

If we have 4.02 moles between the two gases, and we have 1.70 from gas A, then from gas B we simply have:

Total moles = moles A + moles B

moles B = Total moles - moles A

moles B = 4.02 - 1.70

moles B = 2.32 moles

We have 2.32 moles of gas B