JEE Main 2023 · 11 April · morning · Single correct · +4 / −1
Given below are two statements:
Statement-I : Methane and steam passed over a heated Ni catalyst produces hydrogen gas.
Statement-II : Sodium nitrite reacts with NH₄ Cl to give H₂ O, N₂ and NaCl.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (a) Statement I is correct but Statement II is incorrect
- (b) Both the statements I and II are incorrect
- (c) Both the statements I and II are correct
- (d) Statement I is incorrect but Statement II is correct
Answer and solution
Answer (c) Both the statements I and II are correct
Solution
Chemistry ma’am · 3 steps
Check each statement on its own: one is steam reforming, the other is the lab prep of nitrogen.
- Statement I. Methane and steam over a nickel catalyst at about 1270 K is steam reforming. It gives water gas, a mixture of CO and H₂, so hydrogen is formed and the statement is correct.
CH₄ + H₂O →(Ni, 1270 K) CO + 3H₂
- Statement II. Warming aqueous NH₄Cl with NaNO₂ is the standard laboratory preparation of nitrogen: NH₄Cl + NaNO₂ → N₂ + 2H₂O + NaCl. The products are exactly H₂O, N₂ and NaCl, so this statement is also correct.
- So. Both statements are correct, which is option (c).
The idea to keep. Remember the lab prep of N₂: NH₄Cl with NaNO₂, which goes through NH₄NO₂. Steam over hot hydrocarbons on Ni is how hydrogen is made industrially.
Answer from the NTA final answer key. Solution by PrepWiser.