O2 + H2 —> H2O
Balance the equation so that The number of the elements are the equation, and there can not be a fraction
O2 + 2H2 —-> 2 H2O
Since H2 is being consumed 2x as fast as O2, divide the amount of mole of H2 by 2
8.3moles H2/2 is 4.15 moles
Compare the adjusted moles of H2 with the moles of O2.
2.9moles of O2, 4.15 moles of H2
Since there are more moles, after adjustment, of H2, the reaction will run out of O2 and have leftover/unused H2.
The compound that runs out first is the limiting reactant(in this case O2), and the compound that has leftover is the excess reactant(in this case H2)