Question

When silver nitrate is added to an aqueous solution of calcium chloride, a precipitation reaction occurs that removes the chloride ions from solution. $$ 2 \mathrm{AgNO}_{3}(s)+\mathrm{CaCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q) $$ (a) If a solution contains \(10.0 \mathrm{~g} \mathrm{CaCl}_{2}\), what mass of \(\mathrm{AgNO}_{3}\) should be added to remove all of the chloride ions from solution? (b) When enough \(\mathrm{AgNO}_{3}\) is added so that all \(10.0 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) react, what mass of the \(\mathrm{AgCl}\) precipitate should form?

Short answer

a) The required mass of \(\mathrm{AgNO}_{3}\) to remove all the chloride ions from the solution is 30.57 g. b) When enough \(\mathrm{AgNO}_{3}\) is added so that all \(10.0 \, g\) of \(\mathrm{CaCl}_{2}\) react, the mass of the \(\mathrm{AgCl}\) precipitate that should form is 25.8 g.

Step-by-step solution

Find Moles of CaCl2

Using the molar mass of calcium chloride (\(M_{CaCl_2} = 111.0 \, g/mol\)), the number of moles of \(CaCl_{2}\) in the solution can be calculated using the formula:\(n_{CaCl_2}=\frac{m_{CaCl_2}}{M_{CaCl_2}}=\frac{10.0 \, g}{111.0 \, g/mol} \approx 0.09 \, mol\).

Apply Stoichiometry to Find Mass of AgNO3

From the balanced equation, it is seen that the molar ratio of \(\mathrm{AgNO}_{3}\) to \(\mathrm{CaCl}_{2}\) is 2:1. Since we need to find the amount of \(\mathrm{AgNO}_{3}\) required to react with \(0.09 \, mol\) of \(\mathrm{CaCl}_{2}\) completely, this means we need \(2*0.09 \, mol = 0.18 \, mol\) of \(\mathrm{AgNO}_{3}\). So, the mass of \(\mathrm{AgNO}_{3}\) can be calculated with its molar mass (\(M_{AgNO_3} = 169.87 \, g/mol\)):\(m_{AgNO_{3}} = n_{AgNO_{3}} * M_{AgNO_{3}} = 0.18 \, mol * 169.87 \, g/mol = 30.57 \, g\).

Use Stoichiometry to Calculate the Mass of the Precipitate (AgCl)

The mole of precipitate that will form is \(0.18 \, mol\) because from the balanced equation, it is observed that 1 mole of \(\mathrm{CaCl}_2\) produces 2 moles of \(\mathrm{AgCl}\). Hence, given the molar mass of silver chloride (\(M_{AgCl} = 143.32 \, g/mol\)), the mass of the precipitate formed can be found as:\(m_{AgCl} = n_{AgCl} * M_{AgCl} = 0.18 \, mol * 143.32 \, g/mol = 25.8 \, g\).

Key concepts

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in a solution to form an insoluble product, usually known as the precipitate. In the given exercise, when silver nitrate (\(\text{AgNO}_3\)) is added to a calcium chloride (\(\text{CaCl}_2\)) solution, the silver ion (\(\text{Ag}^+\)) interacts with the chloride ion (\(\text{Cl}^-\)) to form silver chloride (\(\text{AgCl}\)), an insoluble solid that precipitates out of the solution.
Key characteristics of a precipitation reaction:

• A solid forms and separates from the solution.
• The reaction involves the exchange of ions between compounds.
• The solid formed is typically the result of attraction between oppositely charged ions.

Understanding precipitation reactions is essential for predicting the results of mixing different ionic compounds in aqueous solutions.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (\(\text{g/mol}\)). It is crucial in stoichiometry for converting between mass and moles, allowing for the understanding of how substances react in exact quantities according to their chemical equations. For calcium chloride (\(\text{CaCl}_2\)), the molar mass is calculated by adding the atomic masses of its components: calcium (Ca) and chlorine (Cl).

• Calcium (Ca): 40.08 g/mol
• Chlorine (Cl): 35.45 g/mol x 2 = 70.90 g/mol
• Total molar mass of \(\text{CaCl}_2\) = 111.0 g/mol

Similarly, the molar mass of silver nitrate (\(\text{AgNO}_3\)) is computed by considering the sum of the atomic masses of silver (Ag), nitrogen (N), and oxygen (O). Calculating molar mass accurately is vital to use in stoichiometric calculations when predicting product amounts or determining reactant quantities.

Silver Nitrate

Silver nitrate, with the chemical formula \(\text{AgNO}_3\), is a versatile compound commonly used in precipitation reactions due to its ability to easily dissolve in water and form precipitates with chloride ions. Some essential properties of silver nitrate include:

• It is highly soluble in water, which makes it suitable for participating in aqueous reactions.
• Silver nitrate is used not only in precipitation reactions but also in qualitative analysis for chloride ions.
• It must be handled with care as it can cause skin stains and is a potential irritant.

In chemistry education, silver nitrate helps illustrate the concept of ionic exchange and precipitation, making it one of the more practical compounds for observing chemical reactions in real-time.

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in stoichiometry, ensuring that the law of conservation of mass is upheld in chemical reactions. For the precipitation reaction given in the exercise:\[2 \text{AgNO}_3 (s) + \text{CaCl}_2 (aq) \longrightarrow 2 \text{AgCl} (s) + \text{Ca(NO}_3\text{)}_2 (aq)\]To balance a chemical equation:

• Count the number of atoms of each element on both sides of the equation.
• Adjust coefficients to have the same number of each kind of atom on both sides.
• Check each element again to ensure balance without altering subscripts of molecules.

Balancing is crucial because it reflects the actual ratios of substances involved and ensures accurate stoichiometric calculations for determining reactants and products in chemical reactions.