Question

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 ?

Answer 1

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.

Answer 2

NO2
here is what i did
write the percent compostion,
divide by their R.A.M,
divide by the smallest number,
approximate.