how much N/5 HCl soluation would required in react completely with 1g of CaCo3
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150ml=.15L
1M/10HCl =.1mol/L
0.15L x 0.1mol/L = 0.015 mol H+
And now the balanced equation:
CaCO3 + 2 H+ --> Ca2+ + H2O + CO2
The molar ratio of H+ to CaCO3 is 2 to 1. Therefore 0.0075 mol CaCO3 must have reacted with the acid.
The molar mass of CaCO3 = 100 g/mol
The grams of CaCO3 that reacted = 0.0075 mol x 100 g/mol = 0.75 grams
The percentage of CaCO3 in the limestone sample is (0.75 g / 1 g) x 100%
which is equal to 75%
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