Solve this

The combustion of ethane ($\ce{C2H6}$) with oxygen ($\ce{O2}$) produces carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$).
The unbalanced equation is: $\ce{C2H6 + O2 -> CO2 + H2O}$
Balancing the equation:
$\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$

Molar mass of $\ce{C2H6}$: $(2 \times 12.01) + (6 \times 1.008) = 24.02 + 6.048 = 30.068 \text{ g/mol}$
Molar mass of $\ce{O2}$: $2 \times 16.00 = 32.00 \text{ g/mol}$
Molar mass of $\ce{CO2}$: $12.01 + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol}$

Given mass of $\ce{C2H6} = 100 \text{ g}$
Given mass of $\ce{O2} = 100 \text{ g}$

Moles of $\ce{C2H6}$:

Moles of $\ce{O2}$:

To find the limiting reactant, divide the moles of each reactant by its stoichiometric coefficient from the balanced equation ($\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$).

For $\ce{C2H6}$:

For $\ce{O2}$:

Since $0.446428 < 1.662895$, $\ce{O2}$ is the limiting reactant.

From the balanced equation, 7 moles of $\ce{O2}$ produce 4 moles of $\ce{CO2}$.

Rounding to three significant figures (based on the given masses of 100 g):

The mass of $\ce{CO2}$ produced is $\boxed{78.6 \text{ g}}$.