Answer:
The mass % of ethane is 29.1 %
Explanation:
Step 1: data given
Mass of gaseous mixture = 9.780 grams
Mixture contains ethane and propane
Moles of oxygen needed = 1.120 moles
Molar mass O2 = 32 g/mol
Molar mass of ethane = 30.07 g/mol
Molar mass of propane = 44.1 g/mol
Step 2: The balanced equations
C2H6 + 7/2 O2 → 2CO2 + 3H2O
C3H8 + 5O2 → 3CO2 + 4H2O
Step 3: Calculate moles
n(C2H6) / n(O2) = 1 / 3.5 = 1/X
m(C2H6) = 3.5x mole * 30.07 g/mol
n(C3H8) /n(O2) = 1/5 = x / (1.12 -x)
m(C3H8) = (1.12 - x) /5 * 44.1 g/mol
3.5 * 30x + 1/5 *44.1 (1.12-x) = 9.780 grams
x = 0.3325 moles
Step 4: Calculate mass ethane
2/7 * 30*0.3325 = 2.85 grams
Step 5: Calculate % of ethane
% ethane = (2.85 grams / 9.78 grams ) * 100%
% ethane = 29.1 %
The mass % of ethane is 29.1 %
