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 followingPercentage of calcium carbonate in the chalk is 80%Number of mmoles of HCl reacted with chalk is 2Number of mmoles of HCl reacted with chalk is 3Number 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+2HCl→CaCl2+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 = 1×10−3×100 = 0.1 g Percentage of calcium carbonate = 0.1×1000.125=80%

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