Question

A gas mixture of 16.0 g of CH4 and 30.0 g of C2H6 are at standard temperature and pressure. What is the partial pressure of CH4 in the mixture?

a) 0.125 atm

b) 0.250 atm

c) 0.500 atm

d) 0.750 atm

e) 1.00 atm

Answer 1

Molar mass of CH4 = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

Molar mass of C2H6 = 2*MM(C) + 6*MM(H)

= 2*12.01 + 6*1.008

= 30.068 g/mol

n(CH4) = mass of CH4/molar mass of CH4

= 16.0/16.042

= 0.9974

n(C2H6) = mass of C2H6/molar mass of C2H6

= 30.0/30.068

= 0.9977

n(CH4),n1 = 0.9974 mol

n(C2H6),n2 = 0.9977 mol

Total number of mol = n1+n2

= 0.9974 + 0.9977

= 1.9951 mol

Partial pressure is:

p(CH4),p1 = (n1*Ptotal)/total mol

= (0.9974 * 1)/1.9951

= 0.500 atm

Answer: c