Enthalpy of reaction between ammonia and oxygen I have to ask this here because I asked literally everybody I know and I still disagree with them.


  Using the information in the table below, calculate the heat of released from the following reaction in $\pu{kJ}$ per $\pu{g}$ of $\ce{NO}.$ (Mw. of $\ce{NO}$ is $\pu{30.01 g/mol})$
  
  $$\ce{4 NH3(g) + 5 O2(g) -&gt; 6 H2O(g) + 4 NO(g)}$$
  
  $$
\begin{array}{lrrr}
\hline
  &amp; \ce{NH3} &amp; \ce{H2O} &amp; \ce{NO} \\
\hline
H_\mathrm{f}^\circ~(\pu{kJ/mol}) &amp; -46.1 &amp; -241.8 &amp; +90.3 \\
\hline
\end{array}
$$
  
  a) $-13.29$
  b) $-8.372$
  c) $-7.541$
  d) $-30.16$
  e) $-5.678$  


I'm convinced the answer is c. 

The enthalpy of this reaction for 4 moles of NO is -905.2, to find the enthalpy for 1 mole of NO we have to divide by 4 to get to -226.3.

Then, to find it per gram of NO we have to divide by 30.01, this gives us 7.541.

Am I right ? If not, what did I do that was wrong? Where's my mistake?

Please don't put this as HW help, I asked multiple websites and went through my working 5-6 times but I still couldn't point out where I went wrong.
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,
$$\frac{Q_{rxn}}{m_{NO}} = \frac{\Delta H_{rxn}}{m_{NO}} = \frac{-905.2\pu{kJ}}{4_{mol, NO}(30.01\pu{g//mol})}=-7.541\pu{kJ//g_{,NO}}$$