which is cheaper :40%HCl at the rate of 50 paise per kg or 80%sulphuric acid at the rate of 25 paise per kg to completely neutralize 7 kg of caustic potash?

Reaction of HCl with NaOH and H2SO4:

HCl + KOH  → KCl + H2O

36.5  56

H2SO4 + 2KOH →  K2SO4 + 2H2O

  98  56 x 2

It can be seen that

56 g of KOH is neutralized by 36.5 g of HCl

Therefore,

7000 g of KOH will be neutralized by 36.5/56 x 7000 = 4562.5 g = 4.56 Kg HCl

So price of 4.56 Kg HCl = 4.56 x 50 paise = 228 paise

And

56 x 2 = 112 g of KOH is neutralized by 98 g of H2SO4

Therefore,

7000 g of KOH will be neutralized by 98/112  X 7000 = 6125 g = 6.12 Kg of H2SO4

So price of 6.1 Kg H2SO4= 6.1 x 25 paise = 152.5 paise

Thus, H2SO4 is cheaper.

  • -9

according to me it will be sulphuric acid it looks a very good question

  • -10
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