So in this task first I think you should use the formula of the: Density/percentage
68.8% divided 26.2%= 2.625954198473282
2.625954198473282 X =
13.12977099236641
Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture<span />
Answer:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
2NO(g) + O₂(g) → 2 NO₂
Explanation:
First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.
3H₂ (g) + N₂(g) → 2NH₃ (g)
Afterwards, ammonia reacts to oxygen, to produce NO and H₂O
The equation for the process will be:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:
2NO(g) + O₂(g) → 2 NO₂
3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)
The answer is 18.23432 grams.
Molar mass of KOH= 56.1056
= 18.234 grams
