Help with Stoichiometry and Volumetric Analysis

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SUMMARY

The discussion focuses on calculating the original mass of calcium carbonate (CaCO3) in a piece of chalk after it reacts with hydrochloric acid (HCl) and is subsequently titrated with sodium hydroxide (NaOH). The balanced chemical equations provided are: CaCO3 + 2HCl → CaCl2 + CO2 + H2O and HCl + NaOH → NaCl + H2O. To find the mass of CaCO3, one must first determine the initial moles of HCl in the 250 mL solution and then calculate the moles of HCl that reacted with NaOH during titration.

PREREQUISITES
  • Understanding of stoichiometry and mole calculations
  • Familiarity with titration techniques and calculations
  • Knowledge of balanced chemical equations
  • Basic concepts of molarity and solution preparation
NEXT STEPS
  • Calculate the initial moles of HCl in 250 mL of 0.293 M solution
  • Determine the moles of HCl that reacted with NaOH using the titration data
  • Use stoichiometry to find the moles of CaCO3 that reacted with HCl
  • Convert moles of CaCO3 to grams using its molar mass (100.09 g/mol)
USEFUL FOR

Chemistry students, educators, and laboratory technicians involved in volumetric analysis and stoichiometry calculations.

kara
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I need some help with this question.

A piece of chalk (mainly calcium carbonate) is placed in 250 mL of 0.293 M HCl. All the CaCO3 reacts, realising CO2 gas, and leaving a clear solution. 50.00 mL of the solution is pipetted into another flask. 87.1 mL of 0.0567 M NaOH is required to titrate the HCl remaining in this 50.00 mL portion. What was the original mass of CaCo3 in the piece of chalk?

So far I've found balanced chemical equations from the question... they are:

1. CaCO3 + 2HCl --> CaCl2 + CO2 + H2O

2. HCl + NaOH --> NaCl + H2O

Where should i go from there, the info given is really overwhelming.
 
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You should be able to determine the moles of HCl that had reacted with the CaCO3, to do this you need to know the original, total, moles of HCl and the amount that remained after the reaction with CaCO3.
 

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