The volume of 0.160 M sulfuric acid necessary to react completely with 75.0 g sodium hydroxide is 0.15L
The reaction of sulphuric acid and sodium hydroxide is expressed as:
[tex]2NaOH + H_2SO_4 ==> Na_2SO_4 + 2H_2O[/tex]
Determine the moles of NaOH present
Molar mass of NaOH = 23 + 16 + 1 = 40g/mole
Given that mass of NaOH is 75.0g
Mole of NaOH = 75/40 = 1.875moles
Determine the number of moles [tex]H_2SO_4[/tex] needed to react with 1.875 moles NaOH
Moles of [tex]H_2SO_4[/tex] = [tex]\frac{1.875}{2} = 0.9375moles of H_2SO_4[/tex]
Recall that volume = Molar mass * number of moles
Volume of [tex]H_2SO_4[/tex] = 0.160 * 0.9375
Volume of [tex]H_2SO_4[/tex] = 0.15L
Hence the volume of 0.160 M sulfuric acid necessary to react completely with 75.0 g sodium hydroxide is 0.15L
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