Enthalpy of the reaction between hydrogen and oxygen
The confusion is partly caused by careless use of quantities and units. The value $H = -484\ \mathrm{kJ}$ denotes an enthalpy. However, what is actually meant is the enthalpy per one mole; i.e. the molar enthalpy $H_{\mathrm m} = -484\ \mathrm{kJ/mol}$.

In the definition of the molar reaction enthalpy, the ‘per mole’ does not refer to any particular substance in the equation. Instead it refers to the entire reaction as a whole. Therefore, the reaction must be specified for which this quantity applies. In this case, the enthalpy of $484\ \mathrm{kJ}$ is released when $2\ \mathrm{mol}$ of hydrogen gas react with $1\ \mathrm{mol}$ of oxygen gas to form $2\ \mathrm{mol}$ of gaseous water: 
$$\ce{2H2(g) + O2(g) -&gt; 2H2O(g)}\qquad\Delta H^\circ = -484\ \mathrm{kJ}$$
(By way of comparison, the corresponding value for liquid water is about $-572\ \mathrm{kJ}$.)

Thus, the molar enthalpy relating to the amount of hydrogen is
$$\frac{\Delta H^\circ}{n(\ce{H2})}=\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}=\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}=-242\ \mathrm{kJ/mol}$$
i.e. the released enthalpy per $1\ \mathrm{mol}$ of hydrogen gas is $242\ \mathrm{kJ}$.

However, the given question is asking about $n(\ce{H2}) = 0.50\ \mathrm{mol}$ of hydrogen (and accordingly $n(\ce{O2}) = 0.25\ \mathrm{mol}$ of oxygen). The corresponding enthalpy is
$$\begin{align}
0.5\ \mathrm{mol}\times\frac{\Delta H^\circ}{n(\ce{H2})}&amp;= 0.5\ \mathrm{mol}\times\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}\\[6pt]
&amp;= 0.5\ \mathrm{mol}\times\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}\\[6pt]
&amp;= 0.5\ \mathrm{mol}\times -242\ \mathrm{kJ/mol}\\[6pt]
&amp;= -121\ \mathrm{kJ}
\end{align}$$