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

1 Answer
Nov 13, 2016

Answer:

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"#