Question

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 1.24 mol of Gas A in the mixture, how many moles of Gas B are present?

Answer #1

Use ideal gas equation to calculate total number of moles of gas

PV = nRT

Where,

P = pressure in atm = 5.00 atm

V = volume in liter = 20.0 liter

n = no. of mole =?

R = gas constant = 0.08206
L**.**atm/mol**.**K

T = temperature in Kelvin = 303 K

We can write above equation as

n = PV/RT

Substitute value

n = (5.00 atm X 20.0 L) / (0.08206 L.atm / mol.K X 303 K) = 4.02 mol

Total mole of gas = 4.02 moles

mole of gas A = 1.24 moles

moles of gas B = total mole of gas - moles of gas A = 4.02 - 1.24 = 2.78 moles

**Moles of gas B = 2.78 moles**

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