# Reactants used for manufacturing of mustard gas [closed]

A compound 'X' which is used in manufacture of mustard gas reacts with (ozone) $$\ce{NH4Cl}$$ to form 'Y' and elemental sulfur ($$\ce{S}$$). The compound 'Y' reacts with $$\ce{NaOH}$$ and $$\ce{Ag}$$ in separate reactions to liberate 'M' and 'K'. The compound 'M' is heated with $$\ce{Cu2O}$$ to give product 'K'. The value of $$\frac{M_K}{\text{Atomicity}_Y}$$ is:

The answer given is $$3.50$$.

Mustard gas is $$\ce{Cl-CH2-CH2-S-CH2-CH2-Cl}$$ and I know that it can be produced by the reaction between $$\ce{S2Cl2}$$ or $$\ce{SCl2}$$ with ethene, however, none of the reactants satisfy the reactions given in the question. What could be the reactants?

• I think the question statement is wrong, as N isn't present in any of the reactants. myPAT adv? – Rahul Verma Jul 9 '20 at 12:04
• @Rahul Verma I am PCB student, but yes, one PCM student sent me this. – Ouch Jul 9 '20 at 12:45
• Welcome to Chemistry! A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. – Martin - マーチン Jul 9 '20 at 14:10
• I think i got the answer, not sure (about one part) so i am not posting an answer. First reaction is S2Cl2 reacting with Ammonium chloride (Not sure of role of O3) to give S4N4 and then S4N4 reacts with Ag to give Ag2S and N2 and S4N4 reacts with OH- to give NH3. Ammonia then reacts with cupric oxide to give nitrogen. So K is dinitrogen and Y is S4N4 and so atomicity would be 8 and mass of K is 28. So answer is 3.5. please Correct if there is a mistake somewhere. – Safdar Faisal Jul 9 '20 at 14:38
• @SafdarFaisal Nice attempt. I would say the question is flawed. Something is not correct. Also bad denotion is used. K can also mean potassium. – Nilay Ghosh Jul 9 '20 at 14:47

In the given question, it is given that 'X' is a compound used in the production of Mustard Gas $$\ce{(Cl-CH2CH2)2S}$$. Therefore 'X' may be $$\ce{S2Cl2}$$ or $$\ce{SCl2}$$ since they react with ethene to give $$\ce{(Cl-CH2CH2)2S}$$.

Since it is given that X reacts with $$\ce{NH4Cl}$$, this shows that X is $$\ce{S2Cl2}$$, since

$$\ce{6S2Cl2 + 4NH4Cl -> S4N4 + S8 + 16HCl}$$

So, we get the compound Y is $$\ce{S4N4}$$

Now onto the next statement which says that 'Y' reacts with $$\ce{NaOH}$$ to release 'M' and reacts with $$\ce{Ag}$$ to release 'K'.

$$\ce{S4N4 + 6OH- + 9H2O -> S2O3^2- + 2S3O6^2- +8NH3}$$

$$\ce{S4N4 + 8Ag -> 4Ag2S +2N2}$$

Therefore, we get that 'M' and 'K' are $$\ce{NH3}$$ and $$\ce{N2}$$.The last statement states that 'M' reacts with $$\ce{Cu2O}$$ to give 'K'. Seeing this we confirm that 'M' is $$\ce{NH3}$$ and 'K' is $$\ce{N2}$$

$$\ce{NH3 + CuO -> N2 + Cu + H2O}$$

Therefore, since Mass of 'K' ($$\ce{N2}$$) is 28 amu and atomicity of 'Y' ($$\ce{S4N4}$$) is eight. And so the answer is $$\frac{28}{8}$$ = 3.5