Stoichiometry and Volumetric Analysis
A piece of chalk (mainly calcium carbonate) is placed in 250. mL of 0.217 M HCl. All the CaCO3 reacts, releasing carbon dioxide gas, and leaving a clear solution. 30.00 mL of the solution is pipetted into another flask. 17.1 mL of 0.0555 M NaOH is required to titrate the HCl remaining in this 30.00-mL portion. What was the original mass of CaCO3 in the piece of chalk?
Balanced Equations:
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)
-First, find the moles of HCl remaining after the solution fully reacted with the chalk.
0.0171L NaOH * (0.0555mol NaOH/1L NaOH) * (1mol HCl/1mol NaOH) = 9.405*10^-4 mol HCl unreacted in 30.00mL of solution
Since the original volume was 250mL:
9.405*10^-4 mol HCl * (0.250L/0.0300L) = 0.007909 mol HCl unreacted in original volume of solution
-Next, find the original number of moles of HCl
0.217M HCl * (0.250L) = 0.05425 mol HCl
-Calculate the difference to find the amount of HCl that reacted with the chalk
0.05425 - 0.007909 = 0.04634 mol HCl
-This is how much HCl was consumed by the chalk.
-Now just use stoichiometry to find the original mass of chalk.
0.04634 mol HCl * (1mol CaCO3/2mol HCl) * (100.0g CaCO3/1mol CaCO3) = 2.32g CaCO3