A mixture containing 2.35 g each of CH4(g), C2H4(g) and C4H10(g) is contained in a 1.50 L flask at a temperature of 25°C.?

(a)
Molar mass of CH₄ = (12.0 + 1.0×4) g/mol = 16.0 g/mol
Molar mass of C₂H₄ = (12.0×2 + 1.0×4) g/mol = 28.0 g/mol
Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol

Volume, V = 1.50 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 25) K = 298 K

Gas law : PV = nRT
Then, P = nRT/V

No. of moles of CH₄, n(CH₄) = (2.35 g) / (16.0 g/mol) = 2.35/16.0 mol
P(CH₄) = (2.35/16.0) × 0.08206 × 298 / 1.50 atm = 2.39 atm

No. of moles of C₂H₄, C₂H₄ = (2.35 g) / (28.0 g/mol) = 2.35/28.0 mol
P(CH₄) = (2.35/28.0) × 0.08206 × 298 / 1.50 atm = 1.37 atm

No. of moles of C₄H₁₀, n(C₄H₁₀) = (2.35 g) / (28.0 g/mol) = 2.35/58.0 mol
P(CH₄) = (2.35/58.0) × 0.08206 × 298 / 1.50 atm = 0.66 atm


(b)
Method 1 :
P(Total) = (2.39 + 1.37 + 0.66) atm = 4.42 atm

Method 2 :
Total number of moles of gases, n(total) = [(2.35/16.0) + (2.35/28.0) + (2.35/58.0)] mol = 0.271 mol
P(Total) = 0.271 × 0.08206 × 298 / 1.50 atm = 4.42 atm

1) You have to calculate the moles of each gas. Here's CH4:

2.35 g / 16.043 g/mol = 0.14648 mol

2) Next, use PV = nRT to calculate the pressure:

(P) (1.50 L) = (0.14648 mol) (0.08206 L atm / mol K) (298 K)

P = 2.388 atm

3) Once you've done the above for the other two gases, add the three pressures for the total pressure inside the container.