Given the equation: 2H2 + O2 → 2H2O, if given 10g H2 gas and 15g O2 gas, what is the limiting reactant?

Start by taking a look at the balanced chemical equation for this reaction

#color(red)(2)"H"_text(2(g]) + "O"_text(2(g]) -> 2"H"_2"O"_text((l])#

Notice that you have a #color(red)(2):1# mole ratio between hydrogen gas and oxygen gas.

This means that, regardless of how many moles of oxygen gas you have, the reaction needs twice as many moles of hydrogen gas in order to proceed.

You know that you start with #"10.0 g"# of hydrogen gas nad #"15.0 g"# of oxygen. To determine how many moles of each you have, use their respective molar masses

#10.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2)/(2.0159color(red)(cancel(color(black)("g")))) = "4.961 moles H"_2#

and

#15.0color(red)(cancel(color(black)("g"))) * ("1 mole O"_2)/(32.0color(red)(cancel(color(black)("g")))) = "0.4688 moles O"_2#

Notice that you have quite a significant difference, way bigger than the required #color(red)(2):1# ratio needed, between how many moles of hydrogen gas and how many moles of oxygen you have.

This means that you're dealing with a limiting reagent. How many moles of oxygen would have been needed to react with all the hydrogen?

#4.961color(red)(cancel(color(black)("moles H"_2))) * ("1 mole O"_2)/(color(red)(2)color(red)(cancel(color(black)("moles H"_2)))) = "2.4805 moles O"_2#

Since you don't have that many moles of oxygen, it follows that oxygen is your limiting reagent, i.e. it will determine how much hydrogen reacts and how much remains in excess.

For example, how many moles of hydrogen gas would react if you have #0.4688# moles of oxygen?

#0.4688color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)" moles H"_2)/(1color(red)(cancel(color(black)("mole O"_2)))) = "0.9376 moles H"_2#

The rest of the hydrogen will be in excess.