Question

find the volume of 0.110 m sulfuric acid necessary to react completely with 77.2 sodium hydroxide

Answer 1

reaction between sulfuric acid(H₂SO₄) and sodium hydroxide(NaOH) is a base and hence it is an acid-base reaction

in which salt is formed usually and water is released

Molar mass of NaOH = 39.997 g/mol

so 77.2 g NaOH = 77.2g/39.997g/mol = 1.93 moles

Lets write the reaction

H₂SO₄ + 2NaOH  → 2H₂O + Na₂SO₄

2 mole NaOH needs 1 mole H₂SO₄ to react completly as can be seen from above reaction equation

for 1 moles NaOH ---- 1/2 moles H₂SO₄

for 1.93 moles of NaOH -----1/2 x 1.93 = 0.965 moles of H₂SO₄ is needed

Moles = molarity x Volume(in litres)

0.965 moles = 0.110M(or moles/L) x V

V = 8.77 L

the voulme is coming very large in scale of normal lab experiments but as unit is not mentioned clearly in your question i assumed it as gram but looking at scale of the experiment it could be 77.2 mg in that case the volume will be 1/1000 L or 8.72 mL