How many grams of water are required to produce 5.50L of hydrogen gas at 25.0 degrees C and 755mmHg?

Answer is: 4.02 grams of water are required.
Chemical reaction: BaH₂ + 2H₂O → Ba(OH)₂ + 2H₂.
Ideal gas law: p·V = n·R·T.
p = 755 mm Hg ÷ 760.0 mmHg / atm = 0.993 atm.
T = 25 + 273.15 = 298.15 K.
V(H₂) = 5.50 L.
R = 0,08206 L·atm/mol·K.
n(H₂) = 0.993 atm · 5.5 L ÷ 0,08206 L·atm/mol·K · 298.15 K.
n(H₂) = 0.223 mol.
From chemical reaction: n(H₂O) : n(H₂) = 1 : 1.
n(H₂O) = 0.223 mol.
m(H₂O) = 0.223 mol · 18 g/mol.
m(H₂O) = 4.02 g.