# Question #814fe

May 23, 2016

This question is not well-proposed, in that sodium metal will react with the water solvent as well.

#### Explanation:

But for sodium hydroxide, $N a O H$, we can write the followig reaction:

$2 N a O H \left(a q\right) + {H}_{2} S {O}_{4} \left(a q\right) \rightarrow N {a}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(a q\right)$

$\text{Moles of sulfuric acid}$ $=$ $0.750 \cdot L \times 6.0 \cdot m o l \cdot {L}^{-} 1$ $=$ $4.50 \cdot m o l$.

Thus we need to add $9.0 \cdot m o l$ sodium hydroxide to neutralize the sulfuric acid. $9.0 \cdot m o l \times 40.0 \cdot g \cdot m o {l}^{-} 1$ $=$ $360 \cdot g$.

Why do we need to add $9$ $m o l$ caustic soda rather than $4.5$ $m o l$?