The answer is #8.3# #g# of #H_2#.
First, start with the balanced chemical equation
#H_(2(g)) + 1/2O_(2(g)) -> H_2O_((g))#
Here are the standard enthalpy of formation values for the reactants and the products:
#H_(2(g))# - #DeltaH_f^@ = 0# #(kJ)/(mol e)#;
#O_(2(g))# - #DeltaH_f^@ = 0# #(kJ)/(mol e)#;
#H_2O_((g))# - #DeltaH_f^@ = -241.8# #(kJ)/(mol e)#.
We know that, for this reaction, the change in enthalpy, #DeltaH#, is #-241.8# #kJ#; we also know that #DeltaH# is defined as the sum of the standard enthalpies of formation (#DeltaH_f^@#) of the products minus the sum of the standard enthalpies of formation (#DeltaH_f^@#) of the reactans - each multiplied by their stoichiometric coefficients.
Therefore, we could write
#DeltaH = 1 mol e * (DeltaH_f^@)_(H_2O) - ( 0.5 mol es * (Delta_f^@)_(O_2) + 1 mol e * (Delta_f^@)_(H_2)) = (DeltaH_f^@)_(H_2O) = -241.8kJ#
This means that the quantity of #H_2# used in the reaction will impact #DeltaH# through the mole-to-mole ratio it has with #H_2O#. So, since we know that #DeltaH = -q#, we can determine that we need a new #DeltaH# of
#DeltaH_(n ew) = -q = -1.00 * 10^3 kJ#,
which is bigger (in absolute terms) than the original #DeltaH#. This means that we have more moles of #H_2# reacting to produce more moles of #H_2O#. Let's set up the equation with #x# moles of #H_2O#, instead of 1:
#DeltaH_(n ew) = (x mol es) * (DeltaH_f^@)_(H_2O) -> x = (DeltaH_(n ew))/(DeltaH_f^@)_(H_2O)#, so
#x = (-1.00 * 10^3 kJ)/(-241.8 kJ) = 4.14# moles of #H_2O# need to be produced in order to release that amount of energy.
Since we know that #H_2#'s molar mass is #2.0 g/(mol)#, and that 1 mole of #H_2# produces 1 mole of #H_2O#, we get
#n_(H_2) = 4.14# moles #-> m_(H_2) = n * molarmass = 4.14 mol es * 2.0 g/(mol) = 8.3g#
