A 0.2027 gram sample of finely powdered limestone (mainly CaCO₃) was dissolved in 50 mL of 0.1035 M HCl. The solution was heated to expel CO₂ produced by the reaction. The remaining HCl was then titrated with 0.1018M of NaOH and it required 16.62 mL. Calculate the percentage of CaCO₃ in the limestone sample? CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Explanation:

CaCO₃     +     2HCl     →      CaCl₂    +    CO₂       +       H₂O

100 gm  +  2000 mL of M HCl

Amount of HCl remaining = .1018 M of 16.62 mL

= .1018 x 16.62 mL of M HCl

= 1.69 mL of M HCl

Initial HCl taken = 50 mL of .1035 M HCl

= 5.175 mL of M HCl

HCl reacted = 5.175 - 1.69 mL of M HCl

= 3.485 mL of  M HCl

2000 mL of M HCl reacts with 100 gm of CaCO₃

3.485 mL of M HCl will react with .174 grams of CaCO₃

percentage of calcium carbonate = .174 x 100 / .2027

= 85.84 %

3

explanation:

the answer is 1 i believe