A 9.780 -g gaseous mixture contains ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\( and propane \)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .$ Complete combustion to form carbon dioxide and water requires 1.120 \(\mathrm{mole}\) of oxygen gas. Calculate the mass percent of ethane in the original mixture.

For the complete combustion of ethane and propane, the balanced chemical equations are: Ethane combustion: $$\mathrm{C}_{2} \mathrm{H}_{6} + \frac{7}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + 3 \mathrm{H}_{2} \mathrm{O}$$ Propane combustion: $$\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O}$$

Let x moles of ethane and y moles of propane be present in the mixture. According to the given data, the moles of oxygen required for complete combustion of both ethane and propane is 1.120 moles. Based on the balanced chemical equations, we can set up the following equation: $$\frac{7}{2}x + 5y = 1.120$$

The given mixture has a mass of 9.780 g. We know the molar masses of ethane and propane, which are 30.07 g/mol and 44.10 g/mol, respectively. If x moles of ethane and y moles of propane are present in the mixture, their total mass can be represented by: $$30.07x + 44.10y = 9.780$$

We now have a system of two equations with two variables: $$\frac{7}{2}x + 5y = 1.120$$ $$30.07x + 44.10y = 9.780$$ Solve this system of equations to find the values of x and y. You may use the method of substitution or elimination. Using substitution, we can solve for x in terms of y in the second equation: $$x = \frac{9.780 - 44.10y}{30.07}$$ Now substitute this expression for x into the first equation: $$\frac{7}{2}\left(\frac{9.780 - 44.10y}{30.07}\right) + 5y = 1.120$$ Solve for y to get: $$y = 0.120$$ To find x, substitute the value of y back into the expression for x: $$x = \frac{9.780 - 44.10\cdot 0.120}{30.07}$$ $$x = 0.220$$ Now we know that the mixture contains 0.220 moles of ethane and 0.120 moles of propane.

To find the mass percent of ethane, first find the mass of ethane in the mixture by multiplying the moles of ethane by its molar mass: Mass of ethane = 0.220 moles * 30.07 g/mol = 6.6154 g Next, divide the mass of ethane by the total mass of the mixture and multiply by 100 to obtain the mass percent: Mass percent of ethane = (6.6154 g / 9.780 g) * 100 = 67.63% Hence, the mass percent of ethane in the original mixture is 67.63%.