Question #74040

The trick here is to realize that when a reaction that involves gaseous products takes place at constant temperature and pressure, the mole ratio that exists between the chemical species that take part in the reaction is equivalent to a volume ratio.

In your case, the balanced chemical equation that describes this reaction looks like this

#color(blue)(2)"H"_ (2(g)) + "O"_ (2(g)) -> color(purple)(2)"H"_ 2"O"_ ((g))#

This tells you that every #1# mole of oxygen gas that takes part in the reaction consumes #color(blue)(2)# moles of hydrogen gas and produces #color(purple)(2)# moles of water.

Under the same conditions for pressure and temperature, you can say that every liter of oxygen gas that takes part in the reaction consumes #color(blue)(2)# liters of hydrogen gas and produces #color(purple)(2)# liters of water vapor.

#overbrace((color(blue)(2)color(white)(.)"moles H"_2)/"1 mole O"_2)^(color(red)("mole ratio")) = overbrace((color(blue)(2)color(white)(.)"L H"_2)/"1 K O"_2)^(color(red)("volume ratio"))#

Now, notice that you don't have enough hydrogen gas to ensure that all the volume of oxygen gas takes part in the reaction.

#2.47 color(red)(cancel(color(black)("L O"_2))) * (color(blue)(2)color(white)(.)"L H"_2)/(1color(red)(cancel(color(black)("L O"_2)))) = "4.94 L H"_2#

Since

#overbrace("2.13 L H"_2)^(color(red)("what you have")) " " < " " overbrace("4.94 L H"_2)^(color(red)("what you need"))#

you can say that hydrogen gas will be the limiting reagent, i.e. it will be completely consumed before all the volume of oxygen gas will get the chance to react.

This means that the reaction will produce

#2.13 color(red)(cancel(color(black)("L H"_2))) * (color(purple)(2)color(white)(.)"L H"_2"O")/(color(blue)(2)color(red)(cancel(color(black)("L H"_2)))) = color(darkgreen)(ul(color(black)("2.13 L H"_2"O")))#

The answer is rounded to three sig figs.