Question

A 0.450 gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3(s) + 2HCl(aq) arrow CaCl2(aq) +H2O(l) +CO2(g) The excess HCL (aq) is titrated by 6.45 mL of 0.125 M NaOH(aq). Calculate the mass percentaageof CaCO3(s) in the sample. Please show steps.

Answer 1

Concepts and reason

Titration is a process where a known concentration solution is used for determining the concentration of unknown solution. Known concentration solution called as titrant while unknown concentrated solution called as analyte.

Fundamentals

Mass percentage: mass percentage represents the purity of the component present in the mixture.

Molarity: molarity is the ratio of the number of moles to the volume of the solution.

Therefore, moles of is used to react with excess present in given reaction.

The balanced chemical equation to the given acid-base titration reaction

According to this balanced chemical equation, 1 mole of reacts with 1mole of

So

Number of moles of needed for moles of = moles.

Number of moles of excess =mol

Number of moles of used to react with

Therefore, moles of is used in the reaction with

Given balanced chemical equation

According to the given balanced chemical equation, 1mole of reacts with 2moles of

Ans:

The mass percentage of the in the sample is 74.4%