Question

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

Answer 1

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

Explanation:

We need (i) a stoichiometrically balanced equation:

`H_2SO_4(aq) + 2NaOH(s) rarr Na_2SO_4(aq) + 2H_2O(l)`

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

And (ii) the stoichiometric quantity of sodium hydroxide:

`"Moles of NaOH "=(66.0*g)/(40.00*g*mol^-1)=1.65*mol`

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