Question

What mass of 90% pure caco3 is required to nutralize 2 litre deci normal solution of HCL

Answer 1

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Answer 2

Answer: The answer is 11.11 gm

Explanation:

Acid taken = 2 l of N/10 - HCl

                   = 2000 ml of N/10 - HCl

                   = (2000/10) ml of N - HCl

                   = 200 ml

For complete neutralization,

->1 gm equivalent of HCl = 1 gm equivalent of CaCO3

->1000 ml of N - HCl = 50 gm of CaCO3 [ Here, Equivalent mass of CaCO3 = 100/2 = 50 gm]

->200 ml of N-HCl = (50/1000 * 200) gm of CaCO3 = 10 gm of CaCO3

But, CaCO3 is 90% pure. so

Mass of 90% pure CaCO3 = 10/90 * 100 = 11.11 gm