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

neutrons are in the middle of the atom and the electrons, which are much smaller.