Using Molar Volume in Gas Stoichiometric Calculations

for #1. 2 H₂(g) + O₂ (g) → 2 H₂O (g) ====> note that 2 moles of H₂ produces 2 moles of H₂O

1.24-L H₂ x ( 1 mol H₂ / 22.4-L H₂) x ( 2 mol H₂O / 2 mol H₂) x (18.0-g H₂O / 1 mol H₂O) = 0.996-g H₂O
#2.steps...
1) convert mass H2O to moles H2O
2) use coefficients of balanced equation to convert to moles O2
3) convert moles O2 to L O2.

solution...
10.5 g H2O x (1 mole H2O / 18.0 g H2O) = 0.583 moles H2O

from the balanced equation, 1 mole O2 2 moles H2O.. therefore...
0.583 moles H2O x (1 mole O2 / 2 moles H2O) = 0.292 moles O2

from here you can either use the specific form of the ideal gas law for STP...
0.292 moles O2 x (22.4 L O2 / mole O2) = 6.54 L O2

or you can use the more general form of the ideal gas law...
V = nRT/P = (0.292 moles) x (0.0821 Latm/moleK) x (273 K) / (1 atm)
V = 6.54 L O2
found this on these websites #2. https://answers.yahoo.com/question/index?qid=20090627131045AAokR6Y #1.https://answers.yahoo.com/question/index?qid=20130423194303AAHrkxE
hope this helps you