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150 mL of `N//10` HCl is required to react completely with 1.0 g of a sample of limestone. Calculate the percentage purity of calcium carbonate. |
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Answer» `150 mL (N)/(10)HCl-=150 mL (N)/(10)CaCO_(3)` `underset("Mol.mass")(CaCO_(3))+underset(2g eq.)(2HCl) to CaCl_(2)+H_(2)O+CO_(2)` Eq. mass of `CaCO_(3)=(40+12+48)/(2)=(100)/(2)=50` Mass of `CaCO_(3)` present in 150 mL `N//10` solution. `[NxxExx(V)/(1000)]=50xx(1)/(10)xx(150)/(1000)=0.75g` Purity `=(0.75)/(1)xx100=75%` |
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