A piece of chalk (mainly calcium carbonate) is placed in 250.mL of 0.223 M HCl.
ID: 690336 • Letter: A
Question
A piece of chalk (mainly calcium carbonate) is placed in 250.mL of 0.223 M HCl.All the CaCO3 reacts, releasing carbon dioxide gas, andleaving a clear solution.
30.00 mL of the solution is pipetted into another flask.
25.1 mL of 0.0506 M NaOH is required to titrate the HCl remainingin this 30.00-mL portion.
What was the original mass of CaCO3 in the piece ofchalk? Thank you! A piece of chalk (mainly calcium carbonate) is placed in 250.mL of 0.223 M HCl.
All the CaCO3 reacts, releasing carbon dioxide gas, andleaving a clear solution.
30.00 mL of the solution is pipetted into another flask.
25.1 mL of 0.0506 M NaOH is required to titrate the HCl remainingin this 30.00-mL portion.
What was the original mass of CaCO3 in the piece ofchalk? Thank you!
Explanation / Answer
CaCO3 + 2 HCl ----> CaCl2 + CO2 + H2O No . of moles of HCl , n = Molarity * Volume in L = 0.223 * 0.25 = 0.05575 moles No . of moles of NaOH , n' = 0.0506 M * 0.0251 L = 0.00127 moles NaOH + HCl ----> NaCl + H2O 1 mole of NaOH requires 1 mole of HCl So no . of moles of HCl reacted in the solution left , n'' =0.00127 moles For 30 mL of the solution no . of moles of HCl left isn" For 250 mL of the solution no. of moles of HCl left = (250 *0.00127 ) / 30 = 0.01058 moles So no. of moles of HCl reacted , n"' = n - 0.01058 = 0.04517 moles By obseving the first reaction 1 mole of CaCO3 reacts with 2moles of HCl So , X moles of CaCO3 reacts with 0.04517 moles of HCl X = ( 0.04517 * 1 ) / 2 = 0.022585 moles No . of moles , N = mass / Molar mass Molar mass of CaCO3 = 40 + 12 + 3 * 16 = 100 g So required mass , m = N * Molar mass = 0.022585 * 100 = 2.2585 gRelated Questions
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