Methane reacts with steam at 1273K in the presence of nickel catalyst to form:

The correct answer is CO and H₂.

**Step 1: Identify the reaction type**
Methane (CH₄) reacting with steam (H₂O) in the presence of a nickel catalyst at a very high temperature (1273 K) is the classic *steam‑reforming* process.

**Step 2: Write the overall balanced equation**
The primary reaction for steam reforming is:

$CH$₄ + H₂O → CO + 3 H₂

This equation shows that one mole of methane reacts with one mole of water to produce one mole of carbon monoxide (CO) and three moles of hydrogen gas (H₂).

**Step 3: Understand why this is the dominant pathway at 1273 K**
- The reaction is highly endothermic; the large temperature supplies the required energy.
- Nickel catalyzes the breaking of the C–H bonds in CH₄ and the O–H bonds in H₂O, facilitating the formation of CO and H₂.
- At 1273 K the equilibrium constant favours the forward direction, so the mixture predominantly contains CO and H₂.

**Step 4: Eliminate the other options**
- *CO and H₂O* would imply no net reaction, which contradicts the high‑temperature, catalyst‑driven process.
- *CO₂ and H₂O* correspond to the water‑gas‑shift or oxidation side reactions, which are secondary and not the primary products under the given conditions.
- *CO₂ and H₂* would result from further oxidation of CO (the water‑gas‑shift reaction: CO + H₂O → CO₂ + H₂), but this is not the main outcome of the initial steam‑reforming step.

**Conclusion**
Under the specified conditions—steam, 1273 K, nickel catalyst—the dominant products of methane reforming are carbon monoxide (CO) and hydrogen gas (H₂).