Question

How many moles of calcium chloride, \(\mathrm{CaCl}_{2}\), are required to precipitate the carbonate ion from \(2.5 \mathrm{~L}\) of a \(0.75 M\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution? (Begin by writing a balanced chemical equation for the reaction.)

Short answer

The number of moles of calcium chloride (\(\mathrm{CaCl}_{2}\)) required to precipitate the carbonate ion from \(2.5 \mathrm{~L}\) of a \(0.75 M\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution is \(1.875 \mathrm{~moles}.\)

Step-by-step solution

Writing the balanced chemical equation

First, write out the balanced chemical equation for the precipitation of carbonate ion by calcium chloride. It is given by \(\mathrm{CaCl}_{2} + \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow 2\mathrm{NaCl} + \mathrm{CaCO}_{3}\).

Finding the number of moles of Na2CO3

Here, the volume of the \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution is given as \(2.5 \mathrm{~L}\) and the molarity is \(0.75 M\). The number of moles of a substance can be found by multiplying the molarity by the volume in litres. Hence, the moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\): \(0.75 M * 2.5 \mathrm{~L} = 1.875 \mathrm{~moles}\)

Determining the moles of CaCl2 required.

From the balanced equation, you can see that one mole of \(\mathrm{CaCl}_{2}\) reacts with one mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to form one mole of \(\mathrm{CaCO}_{3}\) and two moles of \(\mathrm{NaCl}\). Therefore, the moles of \(\mathrm{CaCl}_{2}\) required is equal to the moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). Hence, the moles of \(\mathrm{CaCl}_{2}\) needed is \(1.875 \mathrm{~moles}.\)

Key concepts

Molarity

Molarity is a measure of concentration for a chemical solution, indicating the number of moles of a solute per liter of solution. It is denoted using the unit M (molar), with the formula \[\begin{equation}Molarity (M) = \frac{moles of solute}{liters of solution}\end{equation}\].
This concept is crucial in quantifying the precise amount of a substance involved in a chemical reaction. Students can use molarity to calculate the reacting volumes of solutions or to derive the amount of a reactant needed. In the exercise provided, understanding that a \[\begin{equation}0.75M \end{equation}\]
solution of \[\begin{equation}\mathrm{Na}_{2}CO_{3} \end{equation}\]
contains 0.75 moles of solute per liter simplifies the process of determining the amount of \[\begin{equation}\mathrm{CaCl}_{2} \end{equation}\]
required for the precipitation reaction.

Balanced Chemical Equation

A balanced chemical equation is essential to reflect the law of conservation of mass, ensuring the number of atoms for each element is the same on both sides of the equation. Balancing coefficients are used to achieve this. For instance, in the chemical reaction given in the exercise, \[\begin{equation}\mathrm{CaCl}_{2} + \mathrm{Na}_{2}CO_{3} \rightarrow 2\mathrm{NaCl} + \mathrm{CaCO}_{3}, \end{equation}\]
each side of the equation has one calcium atom, two sodium atoms, three chlorine atoms, and one carbon atom with three oxygen atoms. Understanding how to balance equations is vital because it informs the stoichiometric ratios of reactants and products involved in a chemical reaction. These ratios are used directly in stoichiometry calculations for determining the amount of reactants needed or predicting the yield of products.

Mole Concept

The mole concept ties into several areas of chemistry, including stoichiometry and molarity. A mole is a unit that represents \[\begin{equation}6.022 \times 10^{23} \end{equation}\]
(Avogadro's number) entities of a substance, be it atoms, ions, molecules, or other particles. In the context of the provided exercise, the mole concept is applied to quantify \[\begin{equation}\mathrm{Na}_{2} \mathrm{CO}_{3} \end{equation}\]
and \[\begin{equation}\mathrm{CaCl}_{2} \end{equation}\].
With the molarity given, students need to utilize the molar volume of a solution (22.4 L/mol at STP for gases, different for solutions as in this exercise) to establish the number of moles present. The concept of the mole is crucial because it is the bridge between the macroscopic world (liters, grams) and the atomic or molecular scale.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction where two soluble salts react in solution to form one or more insoluble products, known as precipitates. This reaction is widely used for the separation of ions from a solution. In the scenario provided, \[\begin{equation}\mathrm{CaCl}_{2} \end{equation}\]
reacts with \[\begin{equation}\mathrm{Na}_{2}CO_{3} \end{equation}\]
to form \[\begin{equation}\mathrm{CaCO}_{3}, \end{equation}\]
an insoluble salt, that precipitates out of the solution. The ability to predict and identify precipitation reactions is important for chemists in various fields, including analytical chemistry and waste water treatment. By following the stoichiometry of the balanced equation, students can calculate the specific amounts of each reactant required to completely react, without leaving any excess of the other reactant.