The reaction between the hydrochloric acid and the calcium carbonate is: 2 HCl (
ID: 946193 • Letter: T
Question
The reaction between the hydrochloric acid and the calcium carbonate is:
2 HCl (aq) + CaCO3 (s) ---> CaCl2 (aq) + H2O (l) + CO2 (g)
About 90 mL of water and 10.00 mL of 0.5023 M Hydrochloric acid solution was added to a 1.028 g paper sample. Following our procedure the mixture was stirred and then heated just to a boiling to expel the carbon dioxide. Titration of the excess HCl remaining in the mixture required 16.41 mL (corrected volume) of 0.1055 M NaOH. How many milligrams of calcium carbonate were present in the paper sample, and what is the percent calcium carbonate in this sample?
Mass of CaCO3 ___________________ mg
% CaCO3 _________________ %
Please SHOW all work and use correct number of SIGNIFICANT FIGURES. Thank you!
Explanation / Answer
2 HCl (aq) + CaCO3 (s) ---> CaCl2 (aq) + H2O (l) + CO2 (g)
No of mol of HCl = 0.5023*10/1000 = 0.005023 mol
No of mol of excess HCl = No of mol of NaOH = 16.41/1000*0.1055 = 0.00173 mol
No of mol of HCl recated = 0.005023 - 0.00173 = 0.003293 mol
No of mol of CaCo3 = 0.003293/2 = 0.001645 mol
mass of caco3 = 0.001645*100.0869
= 0.165 grams
= 165 mg
% caco3 = 0.165/1.028*100 = 16.05%
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