"CH_4+H_2O\\to CO+3H_2"
As per the above equation,
"1" mole of methane reacts with "1" mole of water to produce "3" moles of Hydrogen.
Moles of methane in "120\\ g=\\frac{120}{16}=7.5"
Moles of water in "120\\ g=\\frac{120}{18}=6.66"
So, water is a limiting reagent and hydrogen formation will depend on mass of water reacted.
Moles of hydrogen reacted"=3\\times \\frac{120}{18}=20"
Mass of "H_2" produced"=2\\times 20= 40 \\ g"
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