Question

At 1.00 atm and 0 °C, a 5.04 L mixture of methane (CH4) and propane (C3H8) was burned, producing 16.5 g of CO2. What was the mole fraction of each gas in the mixture? Assume complete combustion.

Answer 1

idk

Answer 2

http://openstudy.com/study#/updates/52337ad3e4b0af32a078cc04

Answer 3

there u go: H4 + 2 O2 → CO2 + 2 H2O
C3H8 + 5 O2 → 3 CO2 + 4 H2O

(5.04 L) / (22.414 L/mol) = 0.22485 mol gases total

(15.0 g CO2) / (44.00964 g CO2/mol) × (1/1) = 0.34083 mol C total

Let z be the molar fraction of methane in the original mixture.
Then 1-z is the molar fraction of propane.
Set up the equation for the contribution from each gas to the total moles of C:
(0.22485 mol) × z × (1 C / 1 CH4) + (0.22485 mol) × (1-z) × (3 C / 1 C3H8) =
0.34083 mol C
Solve for z algebraically:
z = 0.742 methane
1-z = 0.258 propane