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

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

idk what the question is about