# Nitrogen and oxygen gases are combined to form nitrogen dioxide gas. All gases are at the same temp and pressure. If 10 L of each gas react, determine the identity and volume of the excess reactant left over?

May 28, 2015

Nitrogen will be your excess reactant.

${N}_{2 \left(g\right)} + \textcolor{red}{2} {O}_{2 \left(g\right)} \to 2 N {O}_{2 \left(g\right)}$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between nitrogen and oxygen. This means that, regardless of how many moles of nitrogen react, you'll need twice as many moles of oxygen for the reaction to take place.

Since all the gases are at the same pressure and temperature, you can write

$P {V}_{\text{nitrogen" = n_"nitrogen}} \cdot R T$, and

$P {V}_{\text{oxygen" = n_"oxygen}} \cdot R T$

You can isolate the constant terms on one side of the equation to get

$\frac{P}{R T} = {n}_{\text{nitrogen"/V_"nitrogen}}$ and $\frac{P}{R T} = {n}_{\text{oxygen"/V_"oxygen}}$

This means that, when the gases are at the same pressure and temperature, the mole ratio becomes the volume ratio.

${n}_{\text{nitrogen"/V_"nitrogen" = n_"oxygen"/V_"oxygen" => n_"nitrogen"/n_"oxygen" = V_"nitrogen"/V_"oxygen}}$

So, regardless of how many liters of nitrogen react, you need twice as many liters of oxygen for the reaction to take place. In order for all the nitrogen to react, you would need

10cancel("liters"N_2) * (color(red)(2)" liters "O_2)/(1cancel("liter"N_2)) = "20 L" ${O}_{2}$

SInce you only have 10 L of oxygen, all the oxygen will be consumed and you'll be left with excess nitrogen. The reaction will only consume

10cancel("liters"O_2) * ("1 liter"N_2)/(color(red)(2)cancel("liters"O_2)) = "5 L" ${N}_{2}$

Therefore, you'll be left with

V_"nitrogen excess" = 10 -5 = color(green)("5 L")