1. combustion
2.decompostioni
3.synthesids
4double replacement
5.single replacement
Answer:
There is 1.6 L of NO produced.
Explanation:
I assume you have an excess of NH3 so that O2 is the limiting reagent.
<u>Step 1:</u> Data given
2.0 liters of oxygen reacts with ammonia
<u>Step 2:</u> 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.
Answer:
0.098 moles
Explanation:
Let y represent the number of moles present
1 mole of Ba(OH)₂ contains 2 moles of OH- ions.
Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.
To get the y moles, we then do cross multiplication
1 mole * y mole = 2 moles * 0.049 mole
y mole = 2 * 0.049 / 1
y mole = 0.098 moles of OH- ions.
1 mole of OH- can neutralize 1 mole of H+
Therefore, 0.098 moles of HNO₃ are present.
