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?

The excess reactant is N2, there will be 5L of N2 left over

Explanation:

Step 1: Data given

Volume of each gas = 10 L

All gases are at the same temperature and pressure.

Step 2: The balaned equation

N2 (g) + 2O2 (g) → 2NO2 (g)

Step 3: Calculate the limiting reactant and excess reactant

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

p*V (N2) = n (N2) * R*T

⇒ P/RT = n (N2) / V (N2)

p*V (O2) = n (O2) * R*T

⇒ P/RT = n (O2) / V (O2)

When the gases are at the same pressure and temperature, the mole ratio becomes the volume ratio:

n (N2) / V (N2) = n (O2) / V (O2)

OR

n (N2) / n (O2) = V (N2) / V (O2)

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

For 10 L of N2 we need 20 L of O2, Since there are only 10L of O2, Oxygen will be the limiting reactant

All of there O2 (10L) will react, with 10/2 = 5L of N2

There will be produce 10 L of NO2 (since mol ratio of O2 and NO2 is the same).

The excess reactant is N2, there will be 5L of N2 left over