How much sulfuric acid containing 90.0% H2SO4 by weight is needed for the production of 1000 kg of concentrated hydrochloric acid containing 42.0% HCl by weight?

Question

Hydrochloric acid was once prepared commercially by heating NaCl with concentrated H2SO4. How much sulfuric acid containing 90.0% H2SO4 by weight is needed for the production of 1000 kg of concentrated hydrochloric acid containing 42.0% HCl by weight?

Answer 1

Amount of pure HCl = 0.42×1000=420kg.

Assumed reaction equation: 2NaCl+H₂SO₄➝2HCl+Na₂SO₄.

Approx atomic weights: Na=23, S=32, H=1, O=16, Cl=35.5.

Molecular weights: HCl=36.5, 2HCl=73, H₂SO₄=2+32+64=98, NaCl=58.5, 2NaCl=117, Na₂SO₄=46+32+64=142.

Reaction masses: 117+98➝73+142=215 for balance.

So 73/215 in the product is pure HCl and 98/215 in the reagents is pure H₂SO₄.

The ratio of the two acids is 73:98=HCl:H₂SO₄.

We know how much pure HCl there is so 73/98=420/x where x is the amount of pure H₂SO₄.

From this x=98×420/73=563.84kg approx.

But the H₂SO₄ solution is 90% pure acid, therefore 563.84 = 0.9y where y is the amount of acid solution.

y=563.84/0.9=626.48kg approx.

Answer 2

We have, according to question 2Nacl + H2So4------Na2SO4+2HCl Writting the molecular weights we get 2 NaCl=2(23+35.5)=117g H2SO4=(2+32+64)=98 g Na2SO4=(46+32+64)=142g 2HCl=2(1+35.5)=73 g By question Impure HCl= 1000kg then Pure HCl= 42% of 1000= 420 kg Then from above eqn we have 98 g of H2SO4= 73 g of HCl Or, 73 g of HCl= 98g of H2S04 So, 420 g of HCl= (98*420)/73 g of H2SO4 =563.836 kg of H2SO4 This is pure H2SO4 but according to question we need to find impure H2S04 then Let y be the impure H2SO4 So, 90% of y = 563.836 kg pure H2SO4 0.9 y = 564.836 kg pure H2SO4 Y = 626.484 kg Thus required impure H2SO4 containing 90% pure H2SO4 = y= 626.484 kg