Question

The mass of 80% pure H2SO4 required to completely neutralise 60g of Na OH is

Answer 1

2 Na OH +  H2 SO4  =>  Na2 SO4 + 2 H2 O



=>So two moles of Na OH ie., 2 * 58.44 gms  reacts with one mole ie., 98 gm of Sulfuric acid.


=>60 gm of hydroxide reacts with mass   m of sulfuric acid

=> m  =  60 * 98 / 116.88  gms

The mass of 80% pure acid is then:  100* m / 80

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