Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? show all of the work used to solve this problem.
unbalanced equation: h2 (g) + o2 (g) h2 o(g)

The reaction of oxygen gas with excess hydrogren would be:

2H2 (g) + O2 (g) --> 2H2O (g)
1 mol of O2 yields 2 mol H2O.
Likewise, 1 L of O2 yields 2L H2O.
so, 8.23 L O2 yields 16.46 L H20

hope this help

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