Question

A mixture of helium and oxygen gases is maintained in a 6.34 L flask at a pressure of 3.23 atm and a temperature of 76 °C. If the gas mixture contains 0.650 grams of helium, the number of grams of oxygen in the mixture is ___ g.

Answer 1

Given volume of flask, V=6.34 L, Pressure, P=3.23 atm, Temperature T=76°C=76+273 K=349 K.

From the ideal gas equation, PV=nRT, where n=total number of moles,

R=gas constant=0.0821 L atm mol^-1 K^-1.

Therefore, n=PV/RT

n=(3.23 atm x 6.34 L)/ (0.0821 L atm mol^-1 K^-1x 349 K)

n=0.7147 mol.

Given mass of helium=0.65 g and its molar mass=4 g/mol.

Therefore moles of He=mass/molar mass=0.65 g/4 g/mol=0.1625 mol.

Total number of moles=moles of He + moles of O2.

Moles of O2= total number of moles - moles of He=0.7147 mol - 0.1625 mol=0.5522 mol O2.

Molar mass of O2=32 g/mol.

Therefore mass of O2 in the mixture=moles x molar mass=0.522 mol x 32 g/mol=17.67 g.

the number of grams of oxygen in the mixture is 17.67 g.

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