In the industrial preparation of HNO3, NH3 is burnt in O2 in the presence of a catalyst, according to the following equation;
4NH3 + 502 - 4NO +6HO
If 260ccm3 of NH3 is burnt completely. what volume of O2 is used up and NO is produced
If all of these are gases you may use a shortcut and use volumes as mols directly.
260 cc NH3 x (4 mols NO/4 mols NH3) = 260 x 1 = ? cc NO.
Do the same with O2 using the coefficients for NH3 and O2.
I don't actually understand, pls explain more on it
OK. The long way. Almost all stoichiometry problems are worked this way.
1. Write and balance the equation, You have done that.
2. There are 22.4 L in a mole of gas (at STP and this probably is not at STP which is why I used the shortcut).
mols NH3 = 250/22,400 = 0.0112
3. Using the coefficients in the balanced equation, convert mols NH3 to mols NO. That is
0.0112 x (4 mols NO/4 mols NH3) = 0.0112
4. Convert 0.0112 mols NO to cc NO (if you want volume) = 0.0112 x 22,400 = 250 cc. If you want grams NO, which the problems asked for and I didn't show in my first response, grams = mols NO x molar mass NO = 0.0112 x 30 = ? grams.
The shortcut way is faster BECAUSE you see step 1 divides by 22,400 to convert to mols and step 4 multiplies by 22,400 to convert to volume. You can time by not doing those two steps.
To determine the volume of O2 used up and NO produced in the given reaction, we need to use the stoichiometry of the balanced equation.
The balanced equation is:
4NH3 + 5O2 -> 4NO + 6H2O
From the equation, we can see that for every 4 moles of NH3 burned, 5 moles of O2 are used up and 4 moles of NO are produced.
Given that 260ccm3 (cubic centimeters) of NH3 is burned completely, we first need to convert this volume from cubic centimeters to moles.
To do this, we need to know the molar volume of NH3, which is the volume occupied by one mole of NH3 at a specific temperature and pressure. Let's assume that the temperature and pressure are standard conditions (0 degrees Celsius and 1 atmosphere), under which the molar volume of any gas is approximately 22.4 liters per mole.
So, 260ccm3 is equivalent to 260/1000 = 0.26 liters.
Now we can convert the volume of NH3 from liters to moles using the molar volume:
0.26 liters * (1 mole / 22.4 liters) = 0.0116 moles of NH3
According to the balanced equation, for every 4 moles of NH3 burned, 5 moles of O2 are used up. Therefore, we can set up a ratio:
0.0116 moles NH3 : X moles O2 = 4 moles NH3 : 5 moles O2
Using this ratio, we can calculate the number of moles of O2 used:
X = (0.0116 moles NH3 * 5 moles O2) / 4 moles NH3 = 0.0145 moles O2
We can now convert the moles of O2 back to volume using the molar volume:
0.0145 moles O2 * 22.4 liters/mole = 0.3248 liters
Therefore, the volume of O2 used up in the reaction is approximately 0.3248 liters.
Similarly, we can calculate the moles of NO produced using the same method.
0.0116 moles NH3 : X moles NO = 4 moles NH3 : 4 moles NO
X = (0.0116 moles NH3 * 4 moles NO) / 4 moles NH3 = 0.0116 moles NO
Converting moles of NO back to volume:
0.0116 moles NO * 22.4 liters/mole = 0.2598 liters
Therefore, the volume of NO produced in the reaction is approximately 0.2598 liters.
Please note that these calculations assume ideal conditions and may not be perfectly accurate in a real-world scenario.