The given question is complete, the complete question is:
To measure the amount of calcium carbonate (CaCO) in a seashell, an analytical chemist crushes a 4.80 g sample of the shell to a fine powder and titrates it to the endpoint with 515. mL of 0.140 M hydrogen chloride (HCl) solution. The balanced chemical equation for the reaction is: 2HCI(a)Co (a) H2Co,(aq) + 2Cl (aq)
What kind of reaction is this?
If you said this was a precipitation reaction, enter the chemical formula of the precipitate.
If you said this was an acid-base reaction, enter the chemical formula of the reactant that is acting as the base.
If you said this was a redox reaction, enter the chemical symbol of the element that is oxidized
Calculate the mass percent of CaCO in the sample. Be sure your answer has the correct number of significant digits.
Answer:
It is an acid-base reaction and the mass percent of CaCo3 is 75.2%.
Explanation:
A chemical reaction in which an insoluble salt produces from two soluble salts is termed as precipitation reaction.
A reaction in which atleast exchange of one proton takes place between the two species is termed as an acid-base reaction.
A reaction in which any of the element has a change in oxidation state is termed as redox reaction.
In the mentioned reaction, there is a transfer of H⁺, no precipitation is forming, and no change in oxidation state taking place, thus, it is an acid-base reaction.
In the acid-base reaction, the base refers to the species that accepts hydrogen ion or proton. In the given case, CO₃²⁻ is accepting H+ ion to become H₂CO₃. Hence, CO₃²⁻ is the base.
In order to calculate mass percent of CaCO₃, first there is a need to find the moles of HCl reacted for a solution,
Moles = Molarity × Volume (L)
Moles = 0.140 mol/L × 515 × 10⁻³ L
Moles = 0.0721 mol
Now from the balanced equation, one mole of CaCO₃ needs two moles of HCl.
So, moles of CaCO₃ reacted will be,
= 1/2 × 0.0721 = 0.03605 mol
The mass of calcium carbonate taking part in reaction will be,
= Moles × Molar mass
= 0.03605 × 100 gram/mole
= 3.6086 gram
Mass% of CaCO₃ = Mass of CaCO₃/Mass of sample × 100
= 3.6086 grams/4.80 grams × 100
Mass % = 75.2%
