The difference between the Enthalpy of Reaction $\Delta_{\mathrm{r}} H$ and the Standard Enthalpy of Reaction $\Delta_{\mathrm{r}} H^{\ominus}$
The standard value is not related to a standard temperature (although Standard Enthalpies of Reaction are often given with respect to the standard temperature - they could have been given for any arbitrary temperature). The term "Standard" refers to standard pressure and the reactants and products being in their standard states. The standard states are usually those of the pure compounds. Furthermore gases are usually assumed to behave like ideal gases. This definition of standard states means that Standard Enthalpies of Reaction are often purely calculational quantities since the underlying reactions are practically not feasible. For example, the reactants usually have to be mixed in order to react with one another, meaning that they are not present in their pure form. Similarly, the products of a reaction are often formed as a mixture and not as pure substances. And real gases certainly don't show the same behavior as ideal gases.
So, enthalpically the difference between the Enthalpy of Reaction and the Standard Enthalpy of Reaction is that you would have to add the Enthalpy of Mixing of the reactants to $\Delta_{\mathrm{r}} H$ and subtract the Enthalpy of Mixing of the products from it in order to get $\Delta_{\mathrm{r}} H^{\ominus}$. Furthermore you would have to account for the enthalpic difference between real and ideal gases if there are any gases involved. Furtunately, those differences are usually quite small compared to the often very big Enthalpies of Reaction so that they can be ignored without introducing a large error. But you have to be mindful of the difference between $\Delta_{\mathrm{r}} H^{\ominus}$ and $\Delta_{\mathrm{r}} H$: for example, if $\Delta_{\mathrm{r}} H^{\ominus}$ is small and the reaction is conducted under very non-ideal conditions with respect to the behavior of gases and/or including liquid mixtures with a very large Enthalpy of Mixing, then you have to expect big differences between the (tabulated) standard values and the measured values.
So, in this sense you would have quite a hard time to calculate the exact Standard Enthalpy of Reaction from your measured $\Delta_{\mathrm{R}} H$. But, under the appropriate conditions your measured value is a good approximation.
How to calculate $\Delta_{\mathrm{r}} H$?
The key to solve your calculational problem is to look at the definition of $\Delta_{\mathrm{r}} H$:
\begin{equation}
\Delta_{\mathrm{r}} H = \biggl(\frac{\partial H}{\partial \xi} \biggl)_{p, T} \ ,
\end{equation}
whereby the extent of reaction $\xi$ is defined as
\begin{equation}
\mathrm{d} \xi = \frac{\mathrm{d}n_{i}}{\nu_{i}}
\end{equation}
with $n_{i}$ and $\nu_{i}$ being the amount and the stoichiometric coefficient of substance $i$, respectively.
Consider the reaction:
\begin{equation}
\ce{\nu_{A} A + \nu_{B} B <=> \nu_{C} C + \nu_{D} D}
\end{equation}
From its definition you see that $\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process when $\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B are converted to $\ce{\nu_{C}}$ moles of C and $\ce{\nu_{D}}$ D (= "conversion of one formula unit").
So, using the differential form of your calorimetric formula (under the assumptions you made) and "dividing" it by $\mathrm{d} \xi$ you get
\begin{equation}
\mathrm{d} H = m C \mathrm{d}T \\
\frac{\partial H}{\partial \xi} = m C \frac{\partial T}{\partial \xi} \\
\Delta_{\mathrm{r}} H = m C \frac{\partial T}{\frac{\partial n_{i}}{\nu_{i}}} = m C \nu_{i} \frac{\partial T}{\partial n_{i}} \ ,
\end{equation}
where $m$ is the mass of the stuff you heat in the calorimeter (assumed to be constant) and $C$ its heat capacity.
Considering finite instead of infinitesimal changes of $n_{i}$ leads to
\begin{equation}
\Delta_{\mathrm{r}} H = m C \nu_{i} \frac{\Delta T}{\Delta n_{i}} \ .
\end{equation}
So, in order to get your Enthalpy of Reaction you simply have to pick one substance from your reaction (it doesn't matter which) and substitute its stoichometric coefficient and the amount that was consumed/produced during the reaction into the equation above. Note that since $\xi$ is always positive the sign of $\Delta n_{i}$ is positive for products and negative for reactants (the same is true for $\nu_{i}$).
Side Note: Standard Enthalpy of Formation $\Delta_{\mathrm{f}} H^{\ominus}$
The Standard Enthalpy of Formation is a special Standard Enthalpy of Reaction for which further conditions are in effect:
- the target compound is to be formed by reacting the (pure) elements
it consists of, whereby each element is expected to be in its most
stable modification for the given temperature.
- The reaction equation is to be formulated in such a way that $1 \,
\text{mol}$ of the compound in question is formed.
