Consider the equation. 2h2(g) + o2(g) > > > > 2h2o(g) which represents the correct mass-volume relationship at stp?
a.) 32.00 g of o2 react with an excess of h2 to produce (2 ´ 22.4) l of h2o.
b.) 2.00 g of h2 react with an excess of o2 to produce (2 ´ 22.4) l of h2o.
c.) 32.00 g of o2 react with an excess of h2 to produce 22.4 l of h2o.
d.) (2 ´ 2.00) g of h2 react with an excess of o2 to produce 22.4 l of h2o.

The equation  2H2(g) + O2(g) ---> 2H2O(g) , gives you the theoretical molar ratios of reactants and products:

2 mol H2(g) : 1 mol O2(g) : 2 mol H2O (g)

You know that at SPT 1 mol of gas is equivalent to 22.4 liter of gas, so you also can write these volumetric ratios:  

2*22.4 liter H2(g) : 22.4 litr O2(g) : 2*22.4 liter H2O (g)

Also , using the molar masses you can state the mass ratios:

2*2* 1 g H2: 2*16 g O2 : 2*18 g H2O

equivalent to 4 g H2 : 32 g O2 : 36 g H2O

When you use the information from the three set of ratios you get to

32.0 g O2 (equivalent to 1 mol and 22.4 liter) react with excess H2 to produce 36 g H2O (equivalent to 2 moles and 2*22.4 liter), which is the option A.

option A. 32 g O2 react with an excess H2 to produce 2*22.4 liter H2O  

Seems cool i guess ttyl