NEET · Chemistry · STD 11 - 9. Hydricarbon
Methane reacts with steam at 1273 K in the presence of nickel catalyst to form :
- A CO and \(H _2 O\)
- B \(CO _2\) and \(H _2\)
- C CO and \(H _2\)
- D \(CO _2\) and \(H _2 O\)
Answer & Solution
Correct Answer
(C) CO and \(H _2\)
Step-by-step Solution
Detailed explanation
(C) CO and \(H _2\)
The reaction of methane with steam at 1273 K in the presence of a nickel catalyst is known as steam reforming of methane. It is an industrial method for the preparation of hydrogen gas.
The balanced chemical equation is:
\(CH_4(g)+H_2 O(g) \xrightarrow[Ni]{1273 K} CO(g)+3 H_2(g)\)
The products formed are carbon monoxide \(( CO )\) and hydrogen \(\left( H _2\right)\).
The reaction of methane with steam at 1273 K in the presence of a nickel catalyst is known as steam reforming of methane. It is an industrial method for the preparation of hydrogen gas.
The balanced chemical equation is:
\(CH_4(g)+H_2 O(g) \xrightarrow[Ni]{1273 K} CO(g)+3 H_2(g)\)
The products formed are carbon monoxide \(( CO )\) and hydrogen \(\left( H _2\right)\).
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