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Question

0.125 g of chalk is treated with 50 mL of 0.1 M of HCl and the excess HCl is required 30 mL of 0.1 M NaOH for complete neutralization. Select the correct statements among the following
  1. Percentage of calcium carbonate in the chalk is 80%
  2. Number of mmoles of HCl reacted with chalk is 2
  3. Number of mmoles of HCl reacted with chalk is 3
  4. Number of mmoles of calcium carbonate is 1 


Solution

The correct options are
A Percentage of calcium carbonate in the chalk is 80%
B Number of mmoles of HCl reacted with chalk is 2
D Number of mmoles of calcium carbonate is 1 
CaCO3+2HClCaCl2+H2O+CO2

Chalk contains calcium carbonate. When it is treated with HCl, it is neutralized and the excess HCl is titrated against sodium hydroxide.  

Initial mmoles of HCl = 50×0.1=5 
mmoles of HCl unreacted = 30×0.1=3 
mmoles of HCl reacted = 5 - 3 = 2
mmoles of CaCO3 present 
= 0.5×mmoles of HCl reacted 
= 0.5×2=1
mass of calcium carbonate = 1×103×100 = 0.1 g
Percentage of calcium carbonate = 0.1×1000.125=80%

 

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