Question

one gram mixture of caco3 and NaCl reacts completely with 100ml of n/10 hcl the % of caco3 in the mixture​

Answer 1

Answer:

Explanation:

=> NaCl can't react with HCl as it is neutral salt and CaCO₃ reacts with HCl and  gives CaCl₂ as a product.

CaCO₃ + 2 HCL → CaCl₂ + H₂O + CO₂

According to this formula, to form 1 mole of CaCl₂, 1 mole of CaCO₃ react with 2 mole of HCl.

normality of HCl = 1/10 N = 0.1N

basicity of HCl = 1. Thus, molarity = 0.1 M

=> Here, volume of HCl solution is 100 ml,

Thus, no of mole of HCl = 0.1M × 100mL = 10 milimole = 0.010 mole

=> Therefore, to complete the above chemical reaction, we required 0.010 mole of HCl and 0.005 mole of CaCO₃.

=> Now, mass of CaCO3 is equal to the product of mole of CaCO3 and molar mass of CaCO3.

∴ mass of CaCO3 = 0.005 × 100

= 0.5g

Therefore, the mass of NaCl in mixture:

= 1g - 0.5g = 0.5g

% of CaCO3 in mixture:

= 0.5/(0.5+0.5) × 100

= 0.5 * 100

= 50%

Answer 2

Answer:answer: option (3) 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 ....

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%