Question

A 9.780 -g gaseous mixture contains ethane (C2H6) and propane (C3H8). Complete combustion to form carbon dioxide and water requires 1.120 moles of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Answer 1

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 %