# Question 11c36

May 30, 2015

You'd need $5.24 \cdot {10}^{6} \text{L}$ of oxygen to burn that much methane under those conditions for temperature and pressure.

The balanced chemical equation for the combustion of methane looks like this

$C {H}_{4 \left(g\right)} + \textcolor{red}{2} {O}_{2 \left(g\right)} \to C {O}_{2 \left(g\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

Notice the $1 : \textcolor{red}{2}$ mole ratio that exists between methane and oxygen. This tells you that, regardless of how many moles of methane react, you need twice as many moles of oxygen for the reaction to take place.

So, the first thing you need to do is figure out how many moles of methane you have. To do that, use the ideal gas law equation

$P V = n R T \implies n = \frac{P V}{R T}$

n_(CH_4) = (120/101.325cancel("atm") * 2.00 * 10^(6)cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * 273.15cancel("K")) = "105750 moles" $C {H}_{4}$

This means that you need at least

105750cancel("moles"CH_4) * (color(red)(2)" moles"O_2)/(1cancel("mole"CH_4)) = "211500 moles"# ${O}_{2}$

in order for all the moles of methane to react.

Now, the oxygen is kept under SATP (Standard Ambient Temperature and Pressure) conditions, which imply a temperature of ${25}^{\circ} \text{C}$ and a pressure of 100.0 kPa.

Once again, use th ideal gas law equation, this time to solve for the volume of oxygen

$P V = n R T \implies V = \frac{n R T}{P}$

${V}_{{O}_{2}} = \left(211500 \cancel{\text{moles") * 0.082(cancel("atm") * "L")/(cancel("mol" * cancel("K"))) * (273.15 + 25)cancel("K"))/(100/101.325cancel("atm}}\right)$

${V}_{{O}_{2}} = \textcolor{g r e e n}{5.24 \cdot {10}^{6} \text{L}}$ $\to$ rounded to three sig figs.