Ammonia (NH3) react with oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O). Write and balance the chemical equation. How many liter of NO are produced when 2.0 liters of oxygen reacts with ammonia?
4NH3 + 5O2 --> 4NO + 6H2O
I assume you have an excess of NH3 so that O2 is the limiting reagent.
When all are gases you can use a shortcut where liters count as mols. So 5 L O2 will produce 4 L NO.
Scale it down to 2 L O2.
To write and balance the chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O), we need to consider the number of atoms of each element on both sides of the equation.
The balanced chemical equation is as follows:
4NH3 + 5O2 -> 4NO + 6H2O
Now, let's determine the number of liters of NO produced when 2.0 liters of oxygen reacts with ammonia.
From the balanced equation, we can deduce the stoichiometric ratio between oxygen and nitrogen monoxide. We can see that 5 moles of oxygen produce 4 moles of nitrogen monoxide.
To calculate the volume of NO produced, we need to consider the volume of oxygen in liters. The measurement of volume follows the stoichiometry of the reaction, meaning if we have 5 moles of oxygen, we will have a volume of 5 liters.
Using the stoichiometric ratio, we can set up a proportion to find the volume of NO produced:
5 liters of O2 / 5 moles of O2 = 2 liters of O2 / x moles of NO
Solving for x, we can find the number of moles of NO produced:
x = (5 liters of O2 * 2 moles of NO) / 5 moles of O2
x = 2 moles of NO
Finally, we can convert the moles of NO to liters using the ideal gas law, assuming standard temperature and pressure (STP):
1 mole of any gas at STP occupies 22.4 liters.
Therefore, 2 moles of NO will occupy:
2 moles of NO * 22.4 liters/mole = 44.8 liters
So when 2.0 liters of oxygen reacts with ammonia, the reaction produces 44.8 liters of NO.