I have a simple lab problem:

$30\ \mathrm{mL}$ of $1\ \mathrm{mol/L}$ $\ce{H2SO4}$ reacts with $50\ \mathrm{mL}$ of $1\ \mathrm{mol/L}$ $\ce{NaOH}$ in a neutralization reaction as per the following equation:

$$\ce{H2SO4 + 2NaOH -> 2H2O + Na2SO4}$$

Initial temperature of reactants was $22.0\ \mathrm{^\circ C}$ and final solution temperature was $30.5\ \mathrm{^\circ C}$.

What I CAN solve is the molar heat of reaction for $\ce{NaOH}$, which I found to be $-56.8\ \mathrm{kJ/mol}$. I found that by using: $\mathrm dHm = Q/n = mc\left(\mathrm dT\right) / n$, where $n$ = moles of $\ce{NaOH}$.

$n = 1\ \mathrm{mol/L} \times 0.05\ \mathrm L = 0.05\ \mathrm{mol}$

Then $\mathrm dHm = \left[ \left( 50\ \mathrm{ml} + 30\ \mathrm{ml} \right) \times 4.19\ \mathrm{J/\left(g\ ^\circ C\right)} \times 8.5\ \mathrm{^\circ C} \right] / 0.05\ \mathrm{mol} = -56.8\ \mathrm{kJ/mol}$ since the reaction is exothermic.

The trickier part is when it asks me what the molar heat of reaction for water is in this neutralization reaction. I know the general enthalpy is just $mc\left(\mathrm dT\right) = 2.84\ \mathrm{kJ}$. But then what do I do with this value to find the molar heat of neutralization for water? I have to divide by the number of mols of water as per the equation above, but I don't know which number to plug in? I need explanation, please.


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