Question

1. a gas mixture is prepared by combining 0.31 mol of methane, ( ch_{4} ), 0.25 mol of ethane, ( c_{2}h_{6} ), and 0.29 mol of propane, ( c_{3}h_{8} ). the total pressure exerted by the three gases is 1.50 atm. what is the partial pressure of ethane in this mixture?

a. 0.25 atm
b. 0.29 atm
c. 0.38 atm
d. 0.44 atm
e. 0.30 atm

Explanation:

Step1: Calculate the total number of moles

The total number of moles \(n_{total}=n_{CH_4}+n_{C_2H_6}+n_{C_3H_8}\)
\(n_{total}=0.31 + 0.25+0.29=0.85\space mol\)

Step2: Calculate the mole fraction of ethane

The mole fraction of ethane \(X_{C_2H_6}=\frac{n_{C_2H_6}}{n_{total}}\)
\(X_{C_2H_6}=\frac{0.25}{0.85}\)

Step3: Calculate the partial pressure of ethane

According to Dalton's law of partial pressures \(P_{C_2H_6}=X_{C_2H_6}\times P_{total}\)
\(P_{C_2H_6}=\frac{0.25}{0.85}\times1.50\space atm\)
\(P_{C_2H_6}=\frac{0.25\times1.50}{0.85}=\frac{0.375}{0.85}\approx0.44\space atm\)

Answer:

D. \(0.44\space atm\)