Calculate the volume of oxygen at step required to burn 7.45 litres of ammonia gas
4NH3 + 3O2 → 2N2 + 6H2O
Looks like 3/4 as much O2 as NH3
4 NH3 + 5 O2 --> 4 NO2 + 6 H2O
at STP (I think you mean) 22.4 liters/mol of any old perfect gas
7.45 liters * 1 mol/22.4liters) = 0.333 orr about 1/3 of a mol of NH3
we need 5 mols of O2 for every 4 of NH3
so
1/3 mol * 5/4 = 0.4167 mols of O2
0.4167 mols * 22.4 liters/mol = 9.33 liters O2
or we could have just said 7.45 * 5/4 liters
I am not sure if you should burn the nitrogen :)
Aaah! I don't think N2 is the product. At temperatures sufficient to make the reaction NH3 + O2 go, it will be high enough to form NO. As soon as NO is formed it will react with more O2 to produce NO2. So I think the logical equation is
4 NH3 + 5 O2 --> 4 NO2 + 6 H2O (but oxygen is not balanced..See corrected equation at the end).
A search on Google lists a litany of products; i.e., N2, N2O, NO & NO2. With normal amounts of O2 I think N2 is not possible. With insufficient amount of O2 the product of N2O is likely. With normal amounts of O2 both NO and NO2 are feasible. Of course with NO2 and over time you will see equilibrium amounts of N2O4 and possibly N2O3 and N2O5 depending upon conditions. With questions in the realm of general chemistry I would go with 4 NH3 + 7 O2 --> 4 NO2 + 6 H2O and for the question it would be 7.45 L NH3 x (7 mols O2/4 mols NH3) = ? L O2 required or
7.45 x 7/4 = ? L.
To calculate the volume of oxygen required to burn a specific amount of ammonia gas, we need to use the balanced chemical equation for the combustion reaction between ammonia (NH3) and oxygen (O2).
The balanced chemical equation for the combustion of ammonia is:
4NH3 + 5O2 → 4NO + 6H2O
From the equation, we can see that 4 moles of NH3 react with 5 moles of O2. To calculate the volume of oxygen required, we need to convert the given volume of ammonia gas (in liters) to moles, and then use the mole ratio from the balanced equation to find the corresponding volume of oxygen.
Here are the steps:
Step 1: Convert 7.45 liters of ammonia gas to moles.
To do this, we need to know the molar volume of ammonia gas at the given conditions (temperature and pressure). Let's assume the molar volume of ammonia gas is 22.4 liters/mol at STP (Standard Temperature and Pressure).
Using the conversion factor:
1 mole NH3 = 22.4 liters NH3 (at STP)
So, the number of moles of ammonia gas is calculated as:
Number of moles of NH3 = 7.45 liters NH3 / 22.4 liters/mol = 0.3325 moles NH3
Step 2: Use the mole ratio from the balanced chemical equation to determine the volume of oxygen.
From the balanced equation, we see that 4 moles of NH3 react with 5 moles of O2.
Using the mole ratio:
5 moles O2 / 4 moles NH3 = x moles O2 / 0.3325 moles NH3
Solving for x, we get:
x = (5/4) * 0.3325 moles O2
x ≈ 0.4156 moles O2
Step 3: Convert the moles of oxygen to volume.
Finally, we can convert the moles of oxygen to volume using the molar volume of oxygen at the given conditions (temperature and pressure). Again, let's assume the molar volume of oxygen is 22.4 liters/mol at STP.
Using the conversion factor:
1 mole O2 = 22.4 liters O2 (at STP)
So, the volume of oxygen required is calculated as:
Volume of O2 = 0.4156 moles O2 * 22.4 liters O2/mol = 9.29 liters O2
Therefore, the volume of oxygen required to burn 7.45 liters of ammonia gas is approximately 9.29 liters.