# Question #5b999

Mar 14, 2017

The volume of the acid is 68 mL.

#### Explanation:

The chemical equation is

$\text{H"_2"SO"_4 + "2LiOH" → "Li"_2"SO"_4 + "2H"_2"O}$

Step 1. Calculate the moles of $\text{LiOH}$

$\text{Moles of LiOH" = 0.225 color(red)(cancel(color(black)("L LiOH"))) × "0.150 mol LiOH"/(1 color(red)(cancel(color(black)("L LiOH")))) = "0.033 75 mol LiOH}$

Step 2. Calculate the moles of ${\text{H"_2"SO}}_{4}$

${\text{Moles of H"_2"SO"_4 = "0.033 75" color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol LiOH")))) = "0.016 88 mol H"_2"SO}}_{4}$

Step 3. Calculate the volume of ${\text{H"_2"SO}}_{4}$

${\text{Volume of H"_2"SO"_4 = "0.016 88" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("1 L H"_2"SO"_4)/(0.25 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.068 L H"_2"SO"_4 = "68 mL H"_2"SO}}_{4}$