in an attempt to establish the formula of an oxide of nitrogen , a known volume of the pure gas was mixed with hydrogen and passed over a catalyst at a suitable temperature. 100% conversion of the oxide to ammonia and water was shown to have taken place.
NxHy gives xNH3 + YH20
2400 cm3 of the nitrogen oxide measured at rtp produced 7.20g of water.The ammonia produced was neutralised by 200 cm3 of 1 mol per dm3.
What was the oxidation number of the nitrogen in the nitrogen oxide ?
NxOy + H2 ==> NH3 + H2O
Here is what I did,
7.2 g H2O = 7.2/18 = 0.4 mol H2O.
200 cc x 1M = 0.2 mol NH3 produced = 0.2 x 17 = 3.4 g NH3. The product have a mass of 7.2 + 3.4 = 10.6 grams.
0.4 mol O in H2O means we had 0.4 mol O in NO2. 0.2 mol NH3 means we had 0.2 mol N in the NxOy.
N(0.2)O(0.4) = NO2 for the empirical formula of the NO compound.
2NO2 + 7H2 ==> 2NH3 + 4H2O
N is +4 in this compound.
Check my work. I usually don't stay up this late.
To determine the oxidation number of nitrogen in the nitrogen oxide (NxHy), we need to consider the overall charge of the compound.
In ammonia (NH3), nitrogen has an oxidation number of -3 because hydrogen has an oxidation number of +1, and the overall charge of ammonia is 0.
Since the nitrogen oxide reacts with hydrogen to produce ammonia, we can assume that the oxidation number of nitrogen in the nitrogen oxide is also -3.
The formula for the reaction is:
NxHy → xNH3 + yH2O
From the given information, we know that 2400 cm3 of the nitrogen oxide produced 7.20g of water. To determine the number of moles of water produced, we can use the formula: n = m/M, where n is the number of moles, m is the mass, and M is the molar mass of water.
Molar mass of water (H2O) = 18.016 g/mol
n = 7.20 g / 18.016 g/mol ≈ 0.3997 mol
Since the ammonia produced was neutralized by 200 cm3 of a 1 mol/dm3 solution, we can determine the number of moles of ammonia produced. The molarity (M) of a solution is defined as the number of moles of solute (ammonia) per liter of solution.
Molarity (M) = Moles of solute (ammonia) / Volume of solution (200 cm3 or 0.2 dm3)
1 mol/dm3 = n / 0.2 dm3
Rearranging the equation to solve for n (moles of ammonia produced):
n = 1 mol/dm3 × 0.2 dm3 = 0.2 mol
Since ammonia (NH3) is produced in a 1:1 ratio with the nitrogen oxide (NxHy), we can equate the number of moles of ammonia produced (0.2 mol) with the number of moles of nitrogen oxide:
x = 0.2 mol
Therefore, the oxidation number of nitrogen (N) in the nitrogen oxide (NxHy) is -3.
To determine the oxidation number of nitrogen in the nitrogen oxide, we need to analyze the chemical reaction given and the information provided.
The balanced equation is:
NxHy -> xNH3 + yH2O
From the equation, we can see that one mole of nitrogen oxide reacts to form x moles of ammonia and y moles of water, where x and y represent the stoichiometric coefficients.
We are given that 2400 cm^3 of the nitrogen oxide measured at rtp (room temperature and pressure) produced 7.20g of water. Since the density of water is known to be 1 g/cm^3, we can convert the mass of water to volume:
Mass of water = 7.20g
Density of water = 1 g/cm^3
Volume of water = Mass of water / Density of water
Volume of water = 7.20 cm^3
Therefore, 2400 cm^3 of nitrogen oxide produces 7.20 cm^3 of water.
Now, we are also given that the ammonia produced was neutralized by 200 cm^3 of a 1 mol/dm^3 solution. This means that one mole of ammonia is formed from 200 cm^3 of the nitrogen oxide.
From the given information, we have:
Volume of nitrogen oxide = 2400 cm^3
Volume of water = 7.20 cm^3
Volume of ammonia = 200 cm^3
Using the volume ratios from the balanced equation, we can set up a relationship:
2400 cm^3 of nitrogen oxide -> 200 cm^3 of ammonia
From the relationship, we can see that the volume ratio of nitrogen oxide to ammonia is 12:1.
Now, let's consider the stoichiometry of the reaction:
NxHy -> xNH3 + yH2O
From the volume ratio, we know that 2400 cm^3 of nitrogen oxide is equivalent to 12 moles of nitrogen oxide.
Therefore, the oxidation number of nitrogen in the nitrogen oxide can be determined by dividing the total number of moles of nitrogen by the total number of moles of nitrogen oxide.
Since we have established that 2400 cm^3 of nitrogen oxide is equivalent to 12 moles, we can calculate the number of moles of nitrogen in the nitrogen oxide:
Moles of nitrogen = Moles of nitrogen oxide / Volume ratio
Moles of nitrogen = 12 moles / 12
Moles of nitrogen = 1 mole
The oxidation number of nitrogen in the nitrogen oxide is determined by the number of electrons it gains or loses when it bonds. In the case of ammonia (NH3), nitrogen has an oxidation number of -3.
Therefore, the oxidation number of nitrogen in the nitrogen oxide is -3.
NO2
here is what i did
write the percent compostion,
divide by their R.A.M,
divide by the smallest number,
approximate.