# When you add 0.857 g of sodium metal to an excess of hydrochloric acid, you find that 8910 J of heat are produced. What is the enthalpy of the reaction?

Sep 11, 2017

#### Answer:

$\Delta {H}_{\text{rxn}}^{\circ} = - 238.9 \cdot k J \cdot m o {l}^{-} 1$

#### Explanation:

We interrogate the reaction.....

$N a \left(s\right) + H C l \left(a q\right) \rightarrow N a C l \left(a q\right) + \frac{1}{2} {H}_{2} + \Delta$

$\text{Moles of sodium} = \frac{0.857 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 0.0373 \cdot m o l$

And thus $\Delta {H}_{\text{rxn}}^{\circ} = \frac{8910 \cdot J}{0.0373 \cdot m o l} = - 238.9 \cdot k J \cdot m o {l}^{-} 1$, and we mean per mole of reaction as written.