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?
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Ответ:
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
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