Enthalpy of reaction between ammonia and oxygen
The heat of reaction of a mixture is defined as the extent of reaction, $\xi$ times the enthaply of reaction, $\Delta H_{rxn}$,
$$Q_{rxn} =\xi\Delta H_{rxn} = \sum^n_{i=1}\nu_iH_{f,i}^{o} =\sum^{n,products}_{i=1}\|\nu_i\|H_{f,i}^{o} - \sum^{m,reactants}_{i=1}\|\nu_i\|H_{f,i}^{o} $$
For a reaction system,
$$\ce{4NH_3(g) + 5O_2(g) -&gt; 6H_2O(g) + 4NO(g)}$$
In the case of the example you've provided, you are assuming this reaction is taking place at standard conditions. If we assume for a stiociometic amount of reactants, then the extend of reaction, $\xi$, should be equal to 1. Therefore,
$$Q_{rxn} = 1*\Delta H_{rxn} = 6\Delta H^o_{f,H_2O} + 4\Delta H^o_{f,NO} - 5\Delta H^o_{f,O_2} - 4\Delta H^o_{f,NH_3}$$
Since, $\Delta H^o_{f,O_2} = 0 \pu{kJ//mol}$, the equation reduces to,
$$\Delta H_{rxn} = 6\pu{mol}(-241.8 \pu{kJ//mol}) + 4\pu{mol}(90.3 \pu{kJ//mol}) - 4\pu{mol}(-46.1 \pu{kJ//mol}) $$
$$\Delta H_{rxn}= -905.2\pu{kJ}$$
Now, per gram of $\ce{NO}$, assuming the basis of 4 mol $\ce{NO}$ from the reaction equation,
$$\frac{Q_{rxn}}{m_{NO}} = \frac{\Delta H_{rxn}}{m_{NO}} = \frac{-905.2\pu{kJ}}{4_{mol, NO}(30.01\pu{g//mol})}=-7.541\pu{kJ//g_{,NO}}$$