Answer:
There is 1.6 L of NO produced.
Step-by-step explanation:
I assume you have an excess of NH3 so that O2 is the limiting reagent.
Step 1: Data given
2.0 liters of oxygen reacts with ammonia
Step 2: The balanced equation
4NH3 + 5O2 → 4NO + 6H2O
For 4 moles of NH3, we need 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O
Consider all gases are kept under the same conditions for pressure and temperature, we can express this mole ratio in terms of the volumes occupied by each gas.
This means: when the reaction consumes 4 liters of ammonia (and 5 liters of oxygen) it produces 4 liters of nitrogen monoxide
Now, when there is 2.0 liters of oxygen consumed, there is 4/2.5 = 1.6 L of nitrogen monoxide produced.
There is 1.6 L of NO produced.