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
