A gaseous mixture of ethane, CH3CH3 and propane, CH3CH2CH3 weighing 1.000 g is burned in excess oxygen, O2(g) to form water, H2O(l) and 1676 mL of carbon dioxide, CO2(g) at 25.0 °C and 100 kPa. What was the percentage by mass of ethane in the original mixture?
- Roger the MoleLv 78 months ago
Let z be the mass (in grams) of ethane.
Then (1.000 - z) is the mass of propane.
2 CH3CH3 + 7 O2 → 4 CO2 + 6 H2O
CH3CH2CH3 + 5 O2 → 3 CO2 + 4 H2O
(z g) / (30.069 g CH3CH3/mol) x (4 mol CO2 / 2 mol CH3CH3) = (0.06651 z) mol CO2 from CH3CH3
(1.000 - z) / (44.0956 g CH3CH2CH3/mol) x (3 mol CO2 / 1 mol CH3CH2CH3) =
(0.068034 - 0.068034 z) mol CO2 from CH3CH2CH3
n = PV / RT = (100 kPa) x (1.676 L) / ((8.3144626 L kPa/K mol) x (25.0 + 273.15) K) =
0.06761 mol CO2 total produced
Add the two expressions for CO2 from each component of the mixture, and set the sum equal to the total of CO2 calculated above, all units in moles:
(0.06651 z) + (0.068034 - 0.068034 z) = 0.06761
Solve for z algebraically:
z = 0.2782 g CH3CH3
(0.2782 g CH3CH3) / (1.000 g total) = 27.8% ethane by mass
- Anonymous8 months ago
Write the balanced equations.
Calculate how many moles of CO2 you have. That gives you moles of C
Let X be the mass of ethane. moles of ethane = X / molar mass
1 g - X = mass of propane. moles of propane = (1g - X ) / molar mass
moles of ethane * 2 + moles of propane * 3 = moles of C from the CO2
Solve for X.