Question

How many grams of sodium hydroxide are required to neutralize 2500.0 mL of 5.493 M of sulfuric acid? A balance equation maybe required.

Answer 1

$2NaOH +H_2SO_4 \to Na_2SO_4+2H_2O$

Thus, 2 moles Sodium hydroxide reacts with 1 mole of Sulfuric acid.

Let x moles react with 2500.0 mL of 5.493 M of sulfuric acid

$\implies x=2.5*5.493/2=6.86625 moles$

Molar mass of NaOH is 40 grams.

Thus, $40*6.86625 =274.65 gms.$ are needed.