Question

150 ml of n/10 HCL are required to react completely with 1.0 grams of a symbol of a limestone caco3 the percentage purity of the sample is​

Answer 1

Reaction:

CaCO3+2HCl⟶CaCl2+H2O+CO2

Number of milli equivalence of HCl used initially is 15

Number of milli moles of HCl used initially is 152

2 moles of HCl reacts with 1 mole of CaCO3.

So amount of calcium carbonate is 154×10−3 mol=38 g

But answer given is 75%

share improve this question

asked

Jan 4 '16 at 7:31

Aditya Dev

3,870●5●37●94 edited

Feb 27 '18 at 13:26

Gaurang Tandon

5,467●6●28●66

1 Answer

order by

up vote

2

down vote

accepted

We know that 2 moles of HCl are required to react completely with 1 mole of CaCO3. We find out the number of moles of HCl in 150ml 0.1M HCl (0.1NHCl=0.1MlHCl), which is 0.015 moles. Now the amount of CaCO3 reacted is (0.0152) = 0.0075 moles.

Amount of CaCO3 in gms =

0.0075 moles×100 (mol.wt of CaCO3)=0.75g

...implying the purity is 75%.