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 ?

Hmm, well it seems like the nitrogen in the nitrogen oxide really had to make a good impression if it got completely converted to ammonia and water. I guess you could say it really hit a "nitr-ogen" when it comes to oxidation!

To determine the oxidation number of nitrogen, let's break down the information. We know that 2400 cm3 of the nitrogen oxide produced 7.20g of water. We also know that the ammonia produced was neutralized by 200 cm3 of 1 mol per dm3.

Now, since we're dealing with a neutralization reaction, we can assume that the number of moles of ammonia is equal to the number of moles of the nitrogen oxide. So, we need to find the number of moles of ammonia.

First, we need to convert the volume of the neutralizing solution from cm3 to dm3. Since 1 dm3 is equal to 1000 cm3, 200 cm3 is equal to 0.2 dm3.

Now, if 1 mol per dm3 of the neutralizing solution is enough to neutralize the ammonia, and we used 0.2 dm3 of the solution, then the number of moles of ammonia produced is 0.2 mol.

Since we have the number of moles of ammonia, we can now find the number of moles of water produced. From the given mass of water (7.20g), we can calculate the number of moles using the molar mass of water (18 g/mol). This gives us 0.4 mol of water.

Now, according to the balanced equation NxHy gives xNH3 + yH2O, we can see that for every mole of nitrogen oxide (NxHy), x moles of ammonia (NH3) and y moles of water (H2O) are produced.

So, in this case, we have x = 0.2 mol of ammonia and y = 0.4 mol of water. Therefore, the ratio of x to y is 0.2:0.4, which simplifies to 1:2.

Given that the formula of the nitrogen oxide is NxHy, we can now determine the ratio of x to y. Since x: y = 1:2, it means that the oxidation number of nitrogen in the nitrogen oxide is +2.

Well, there you have it! The nitrogen in the nitrogen oxide sure has a good sense of humor, with an oxidation number of +2. It knows how to "nitro-gen-erate" some laughs!

To determine the oxidation number of nitrogen in the nitrogen oxide, we need to use the information provided in the question.

1. From the given balanced equation: NxHy gives xNH3 + yH2O, we know that the nitrogen oxide is being converted to ammonia (NH3) and water (H2O).

2. We are told that 2400 cm3 of the nitrogen oxide produced 7.20g of water. We can use this information to determine the number of moles of water produced.

- First, we need to convert the volume of the nitrogen oxide to moles. Since we measured the gas at rtp (room temperature and pressure), we can use the ideal gas law equation: PV = nRT.

- From the ideal gas law, we can rearrange the equation to solve for moles:
n = PV/RT

- Plugging in the values:
n = (2400 cm3)(1 dm3/1000 cm3)(1 mol/dm3) / (0.0821 mol·K-1·dm3·K-1 × 298 K)
n ≈ 0.0997 mol

- The number of moles of water produced is equal to the moles of nitrogen oxide since the balanced equation shows a 1:1 stoichiometric relationship. Therefore, the moles of water produced is also 0.0997 mol.

3. We are also given that the ammonia produced was neutralized by 200 cm3 of 1 mol per dm3. We can use this information to determine the number of moles of ammonia produced.

- First, we convert the volume of ammonia to moles, using the same approach as in step 2:
n = (200 cm3)(1 dm3/1000 cm3)(1 mol/dm3)
n = 0.2 mol

- Since the balanced equation shows a stoichiometric ratio of 1:1 between ammonia and nitrogen oxide, the moles of ammonia produced is also 0.2 mol.

4. From step 3, we have determined that the moles of ammonia produced are equal to the moles of nitrogen oxide present. Therefore, the moles of nitrogen oxide are also 0.2 mol.

5. We are given the mass of water produced (7.20g) and we know the molar mass of water (H2O) is approximately 18.02 g/mol.

- We can calculate the moles of water produced:
n = mass / molar mass = 7.20g / 18.02 g/mol ≈ 0.3996 mol

6. Since the balanced equation shows that the ratio of water to nitrogen oxide is 1:1, we can equate the moles of water produced (0.3996 mol) to the moles of nitrogen oxide:
(0.3996 mol of H2O) = (0.0997 mol of NxHy)

7. Solving for x and y in NxHy:
x = 0.0997 mol
y = 0.2003 mol

8. The oxidation number of nitrogen in the nitrogen oxide can be determined using the formula:
Oxidation number = x - 3y

Plugging in the values:
Oxidation number = 0.0997 - (3 × 0.2003)
= 0.0997 - 0.6009
≈ -0.5012

Therefore, the oxidation number of nitrogen in the nitrogen oxide is approximately -0.5012.

To determine the oxidation number of nitrogen in the nitrogen oxide, we can follow the steps provided in the question. Let's break it down:

Step 1: Convert 2400 cm^3 of nitrogen oxide to moles.
To do this, we need to know the molar volume of a gas at room temperature and pressure (rtp). The molar volume at rtp is approximately 24 dm^3. Therefore, we can convert cm^3 to dm^3 by dividing by 1000:
2400 cm^3 = 2400/1000 = 2.4 dm^3

Now, we can use the molar volume to convert to moles:
Moles of nitrogen oxide = Volume (in dm^3) / Molar volume
Moles of nitrogen oxide = 2.4 dm^3 / 24 dm^3 = 0.1 moles

Step 2: Determine the moles of water produced.
From the question, we know that 7.20 g of water was produced. We can convert grams to moles using the molar mass of water:
Molar mass of water (H2O) = (2 × Molar mass of hydrogen) + (1 × Molar mass of oxygen)
Molar mass of hydrogen = 1 g/mol
Molar mass of oxygen = 16 g/mol
Molar mass of water = (2 × 1 g/mol) + (1 × 16 g/mol) = 18 g/mol

Moles of water = Mass of water / Molar mass of water
Moles of water = 7.20 g / 18 g/mol = 0.4 moles

Step 3: Determine the ratio of moles of nitrogen oxide to moles of water.
From the balanced equation given in the question: NxHy gives xNH3 + yH2O

Comparing the coefficients, we can see that the ratio of moles of nitrogen oxide to moles of water is 1:Y.

Step 4: Determine the value of Y in the ratio of moles.
Since 0.1 moles of nitrogen oxide produces 0.4 moles of water, the value of Y can be calculated by dividing the moles of water by the moles of nitrogen oxide:
Y = Moles of water / Moles of nitrogen oxide = 0.4 moles / 0.1 moles = 4

So, in the formula NxHy, the value of Y is 4. Therefore, the oxidation number of the nitrogen in the nitrogen oxide is +4.