!! LONG ANSWER !!
Here's how you'd go about solving this problem.
Your reaction will initially form liquid water at
- Heating the water from
- Convert the water from liquid at
#100^@"C"#to vapor at #100^@"C"#- this represents a phase change.
- Heat the water vapor from
#100^@"C"#to whatever the final temperature will be.
So, start with the balanced chemical equation for the formation of liquid water
You know that the standard enthalpy of formation for water is
SIDE NOTE The standard enthalpy of formation is defined at 25 degrees Celsius, but since your temperature is very close to that, I'll assume it to be equal to the value measured at standard conditions.
However, your reaction doesn't produce 1 mole, it actually produces 6 moles of water. Because hydrogen gas and water have a
So, 6 moles of hydrogen gas react
Since you have 6 moles of water, you'll have
So, to heat water from
To go from liquid at
From this point on, you'll use the heat given off by the formation reaction to heat the vapor from
The remaining amount of enery will be
Now solve for the final temperature of the water vapor
Rounded to two sig figs, the number of sig figs given for 6.0 and 3.0 moles, the answer will be