- #1

skepticwulf

- 74

- 1

## Homework Statement

A 9.780-g gaseous mixture contains ethane (C2H6) and propane

(C3H8). Complete combustion to form carbon dioxide

and water requires 1.120 mole of oxygen gas. Calculate the

mass percent of ethane in the original mixture.

## Homework Equations

## The Attempt at a Solution

7/2 mol O2 burns 1 mole ethane, then x mole O2 burns 2x/7 mole ethane (A)

5 mole O2 burns 1 mole propane, then (1.12-x) mole O2 burns (1.12-x)/5 mole propane (B)

Multiplying A and B with ethane's and propane's molar masses and adding them will give 9,78g.

So solving that for x and then recalculating A for ethane and finally then dividing by 9.78 should give mass percent of ethane.

But, this solution disagrees with solution manual, mine is %29,14, textbook's %38.

I

**do**understand the solution(there's a solution besides an answer) of textbook but I

**do not**understand why mine is incorrect!