Question

A cheap source of calcium ion in solid form is limestone (CaCO_3). How much of this compound is required to make 2.00 grams of pure CaCl_2?

a) consider the two sources of chloride in the reactant list available for the lab- NaCl and HCl. Write the possible reactions with CaCO_3 in the form of balanced chemical equations. Which one of the two is appropriate for making CaCl and why?

b) Once you have settled on the reactants calculate the amount of the second reactant needed to convert all the calcium carbonate to calcium chloride. How can you be sure that you have added enough to convert all the calcium carbonate? If you have added in excess of the second reagent how will you figure out the amount of excess.

c) Based on the amount of HCl added, what is the calculated yield of CaCl_2 in the reaction?

Answer 1

There is one mol of Ca in CaCO₃ and also one mol in CaCl₂, so we know that we need one mol of carbonate to obtain one mol of chloride. The number of moles in 2.00 grams of CaCl₂ is:

We need 0.0180 moles of CaCO₃, which represent a mass of:

a)

HCl is better, since the subproducts are a gas (it can easily be eliminated) and water (the solvent). NaCl also yields sodium carbonate, which is soluble in water and complicates the separation of CaCl₂.

b) Each mol of CaCO₃ reacts with two moles of HCl, so the needed HCl is:

If CO₂ is released we see bubbling, which means the reaction is happening. Once all the carbonate has reacted, there is no more bubbling. The excess amount of HCl can be measured by measuring the pH of the solution or by titration with a base.

c) To answer this, I need information on the amount added (I suspect this comes from an experimental procedure).