The unbalanced reactions given in List-I give the products in List-II. Match List-I with List-II and give your answer as abcd, where a, b, c and d are the matches of P, Q, R and S respectively (for ex

Question

The unbalanced reactions given in List-I give the products in List-II. Match List-I with List-II and give your answer as abcd, where a, b, c and d are the matches of P, Q, R and S respectively (for example, if P → 2, Q → 3, R → 1 and S → 4, then your answer is 2314). Use the numerical values of the lower-roman labels: i=1, ii=2, iii=3, iv=4.

List-I	List-II
(P) $NaN O_{3} (s) + Na (s) Δ$	(i) $N a_{2} S O_{4}$
(Q) $N a_{2} O_{2} + S O_{2} \to$	(ii) $C_{2} H_{2}$
(R) $K O_{2} + H_{2} O \to$	(iii) $N a_{2} O$
(S) $Ca C_{2} + H_{2} O \to$	(iv) $O_{2}$

TestHub · Accepted Answer

P: 2NaN O 3 ​ + 10Na Δ ​ 6N a 2 ​ O + N 2 ​ , hence P-iii. Q: N a 2 ​ O 2 ​ + S O 2 ​ → N a 2 ​ S O 4 ​ , hence Q-i. R: 2K O 2 ​ + 2 H 2 ​ O → 2KOH + H 2 ​ O 2 ​ + O 2 ​ , hence R-iv. S: Calcium carbide gives acetylene and calcium hydroxide on hydrolysis, hence S-ii. Thus P-iii; Q-i; R-iv; S-ii gives the numerical code 3142.

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