Question

One gram mixture of caco3 and nacl react completely with hundred ml of invite in n/10 hcl the percentage of caco3 in the mixture is

Answer 1

Answer:

Explanation:

=> We can write the chemical reaction as follow for the given question:

CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

=> 1 mol CaCO₃ = 2 mol HCl

moles of HCl = 100 mL * 0.1 N

= 100 / 1000 * 0.1

= 0.01 moles

=> Moles of CaCO₃ = 0.01 / 2 = 0.005 moles

Mass of CaCO₃ = 0.005 mol * 100 g/ml

= 0.5 gram

=> Now, in 1 gm of mixture of NaCl and CaCO₃

mass of NaCl = 1 - 0.5 = 0.5 g

% of NaCl = 0.5 g / 1 g * 100

= 50%

Thus, the percentage of caco3 in the mixture is 50%