Question

A closed container has a mixture of methane, CH4, ethane, C2H6, and propane, C3H8. If 17 g of methane, 21.6 g of ethane, and 30.3 g of propane are placed in the 4.74 L container at 63.7°C, what is the total pressure in the container (in atm)? Assume the gases do not react with each other

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

Molar mass of C3H8 = 3*MM(C) + 8*MM(H)

= 3*12.01 + 8*1.008

= 44.094 g/mol

n(CH4) = mass/molar mass

= 17.0/16.042

= 1.0597

n(C2H6) = mass/molar mass

= 21.6/30.067999999999998

= 0.7184

n(C3H8) = mass/molar mass

= 30.3/44.094

= 0.6872

n(CH4),n1 = 1.0597 mol

n(C2H6),n2 = 0.7184 mol

n(C3H8),n3 = 0.6872 mol

Total number of mol = n1+n2+n3

= 1.0597 + 0.7184 + 0.6872

= 2.4653 mol

we have:

V = 4.74 L

n = 2.4653 mol

T = 63.7 oC

= (63.7+273) K

= 336.7 K

we have below equation to be used:

P * V = n*R*T

P * 4.74 L = 2.4653 mol* 0.0821 atm.L/mol.K * 336.7 K

P = 14.4 atm

Answer: 14.4 atm