Question

Calculate the mass of 60% pure sample of H2SO4 required to neutralize 500 grams of NaOH.

Answer 1

Answer:

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.

500 gm of hydroxide reacts with mass   m of sulfuric acid

      m  =  500 * 98 / 116.88  gms

     m = 422.41

The mass of 80% pure acid is then:  100* 422.41 / 80

so, the mass of the 80 % pure acid is 528.01 g

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