# Given the equation C_2H_6(g) + O_2(g) -> CO_2 (g) + H_2O(g) (not balanced), what is the number of liters of CO_2 formed at STP when 240.0 grams of C_2H_6 is burned in excess oxygen gas?

Jan 21, 2016

$\text{362.4 L}$

#### Explanation:

The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version

$\textcolor{red}{2} {\text{C"_2"H"_text(6(g]) + 7"O"_text(2(g]) -> color(purple)(4)"CO"_text(2(g]) + 6"H"_2"O}}_{\textrm{\left(g\right]}}$

Now, the problem provides you with a mass of ethane, ${\text{C"_2"H}}_{6}$, and asks for the volume of carbon dioxide, ${\text{CO}}_{2}$, formed by the reaction at STP, Standard Temperature and Pressure.

As you know, STP conditions are defined as a temperature of ${0}^{\circ} \text{C}$ and a pressure of $\text{100 kPa}$. Under these specific conditions, one mole of any ideal gas occupies exactly $\text{22.7 L}$ - this is known as the molar volume of a gas at STP.

So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.

Use the molar mass of ethane to determine how many moles you'd get in that $\text{240.0-g}$ sample

240.0 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "7.9814 moles C"_2"H"_6

Now, notice that you have a $\textcolor{red}{2} : \textcolor{p u r p \le}{4}$ mole ratio between ethane and carbon dioxide. Since ethane reacts with excess oxygen, you can assume that all the moles will take part in the reaction.

This means that the reaction will produce

7.9814 color(red)(cancel(color(black)("moles C"_2"H"_6))) * (color(purple)(4)" moles CO"_2)/(color(red)(2)color(red)(cancel(color(black)("moles C"_2"H"_6)))) = "15.9628 moles CO"_2

So, if one mole of any ideal gas occupies $\text{22.7 L}$ at STP, this many moles will occupy

15.9628color(red)(cancel(color(black)("moles CO"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)("362.4 L")

The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.