# What volume of 0.160*mol*L^-1 "sulfuric acid" is required to neutralize a 66.0*g mass of "sodium hydroxide"?

Nov 13, 2016

A bit over $5$ $L$ of the sulfuric acid solution.

#### Explanation:

We need (i) a stoichiometrically balanced equation:

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

And this shows a 1:2 equivalence between the sulfuric acid and the sodium hydroxide.

And (ii) the stoichiometric quantity of sodium hydroxide:

$\text{Moles of NaOH } = \frac{66.0 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} = 1.65 \cdot m o l$

Given the stoichiometry, we need 1/2xx(1.65*mol)/(0.160*mol*L^-1)=??L," sulfuric acid solution"