Question

1g mixture of CaCO3 and NaCl reacts Completly with 100 ml of N/10 HCl. The% of CaCO3 in a mixture is​

Answer 1

50%

explanation:NaCl is neutral salt so, it can't react with HCl but CaCO3 reacts with HCl and forms CaCl2 compound.

To understand, see chemical reaction ....

$caco _{3} + 2hcl = > cacl _{2} + h_{2} o + co_{2}$


here it is clear that , one mole of CaCO3 reacts with 2 moles of HCl.

given, normality of HCl = 0.1N so, molarity = 0.1M [ as basicity of HCl = 1]

volume of HCl solution = 120ml

so, number of mole of HCl = 0.1M × 100mL = 10 milimole or 0.010 mole

so, for 0.010 mole of HCl 0.005 mole of CaCO3 is require to complete chemical reaction.

hence, mass of CaCO3 = mole of CaCO3 × molar mass of CaCO3

= 0.005 × 100

= 0.5g

so, the mass of NaCl in mixture = 1g - 0.5g = 0.5g

now, % of CaCO3 in mixture = 0.5/(0.5+0.5) × 100 = 50%