Answer
Methane is the limiting reactant.
$375\ g\ H_2$
$1392.5\ g\ H_2O$
Work Step by Step
Number of moles of the reactants:
Methane: $995\ g\div 16.04\ g/mol=62.03\ mol$
Water: $2510\ g\div 18.015\ g/mol=139.33\ mol$
Ratio: $139.33/62.03=2.25$
Methane is the limiting reactant.
From stoichiometry:
$62.03\ mol\times 3_{H_2}/1_{CH_4} \times 2.016\ g/mol\ H_2=375\ g\ H_2$
Water reacted:
$62.03\ mol\times 1_{H_2O}/1_{CH_4} \times 18.015\ g/mol\ H_2O=1117.5\ g\ H_2O$
Remaining: $2510-1117.5=1392.5\ g\ H_2O$
