Problem 14 - Chemistry and Chemical Reactivity

Question

The formation of water-insoluble silver chloride is useful in the analysis of chloride-containing substances. Consider the following unbalanced equation: \(\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) (a) Write the balanced equation. (b) What mass AgNO \(_{3},\) in grams, is required for complete reaction with \(0.156 \mathrm{g}\) of \(\mathrm{BaCl}_{2} ?\) What mass of \(\mathrm{AgCl}\) is produced?

Step-by-Step Solution

Answer

The balanced equation is \(\mathrm{BaCl}_2 + 2\mathrm{AgNO}_3 \rightarrow 2\mathrm{AgCl} + \mathrm{Ba(NO}_3\mathrm{)}_2\). 0.2546 g of \(\mathrm{AgNO}_3\) are needed, and 0.2148 g of \(\mathrm{AgCl}\) is produced.

Step 1: Balance the Equation

Identify elements that need balancing. In our reaction, we have 1 barium, 2 chlorides, 1 silver, and 2 nitrates on both sides. Once these elements are equal on both sides, our balanced equation is: \[\mathrm{BaCl}_{2}(\mathrm{aq}) + 2\mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow 2\mathrm{AgCl}(\mathrm{s}) + \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\]

Step 2: Calculate Molar Mass of Reactants and Products

Calculate the molar mass for each compound involved. - \(\mathrm{BaCl}_2\): Ba = 137.33, Cl = 35.45; Molar mass = 137.33 + (2 × 35.45) = 208.23 g/mol.- \(\mathrm{AgNO}_3\): Ag = 107.87, N = 14.01, O = 16.00; Molar mass = 107.87 + 14.01 + (3 × 16.00) = 169.87 g/mol.- \(\mathrm{AgCl}\): Ag = 107.87, Cl = 35.45; Molar mass = 107.87 + 35.45 = 143.32 g/mol.

Step 3: Determine Moles of BaCl2

Using the molar mass of \(\mathrm{BaCl}_2\), convert 0.156 g to moles. - Moles of \(\mathrm{BaCl}_2\) = \(\frac{0.156\, \text{g}}{208.23\, \text{g/mol}}\) = 0.000749 moles.

Step 4: Calculate Moles and Mass of AgNO3 Required

Using the balanced equation, 1 mole of \(\mathrm{BaCl}_2\) reacts with 2 moles of \(\mathrm{AgNO}_3\). Therefore, multiply moles of \(\mathrm{BaCl}_2\) by 2.- Moles of \(\mathrm{AgNO}_3\) required = 2 × 0.000749 moles = 0.001498 moles.Convert moles to grams:- Mass of \(\mathrm{AgNO}_3\) = 0.001498 moles × 169.87 g/mol = 0.2546 g.

Step 5: Calculate Mass of AgCl Produced

According to the balanced equation, 1 mole of \(\mathrm{BaCl}_2\) produces 2 moles of \(\mathrm{AgCl}\). Thus, use the same moles calculated for \(\mathrm{AgNO}_3\) because they react in a 1:1 ratio to form \(\mathrm{AgCl}\).- Moles of \(\mathrm{AgCl}\) produced = 0.001498 moles.Convert moles to grams:- Mass of \(\mathrm{AgCl}\) = 0.001498 moles × 143.32 g/mol = 0.2148 g.

Key Concepts

Chemical Equations

Chemical equations are fundamental in chemistry. They represent the reactants and products in a chemical reaction. Understanding how to balance them is vital. Balancing ensures that the law of conservation of mass is upheld. This means the number of atoms for each element must remain the same on both sides. In the example, we have the reaction \( \mathrm{BaCl}_{2}(\mathrm{aq}) + \mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s}) + \mathrm{Ba}(\mathrm{NO}_{3})_{2}(\mathrm{aq}) \).
The process of balancing begins with adding coefficients to compounds where needed. Each element should have an equal count on both sides.

• Barium (Ba): 1 on each side.
• Chloride (Cl): 2 on the left; hence, needs to match 2 AgCl on the right.
• Silver (Ag): Balanced with 2 on each side.
• Nitrate (NO\(_3\)): Balanced with 2 on each side.

As a result, the balanced chemical equation is: \[ \mathrm{BaCl}_{2}(\mathrm{aq}) + 2\mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow 2\mathrm{AgCl}(\mathrm{s}) + \mathrm{Ba}(\mathrm{NO}_{3})_{2}(\mathrm{aq}). \] Proper balancing guides the stoichiometric calculations needed in reactions.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance. It is a crucial concept when solving stoichiometry problems. To calculate it, sum the atomic masses of all atoms in the formula. This is crucial when converting between grams and moles.
For \( \mathrm{BaCl}_2 \): Add the masses of Barium (137.33) and Chloride (2 \times 35.45) to get 208.23 g/mol.
For \( \mathrm{AgNO}_3 \): Combine Silver (107.87), Nitrogen (14.01), and Oxygen (3 \times 16.00) to find 169.87 g/mol.
For \( \mathrm{AgCl} \): Silver (107.87) plus Chloride (35.45) equals 143.32 g/mol.

• Conversion example: Calculate moles of \( \mathrm{BaCl}_2 \) from 0.156 g using\[ \frac{0.156 \, \text{g}}{208.23 \, \text{g/mol}} \approx 0.000749 \, \text{moles}. \]

The molar mass calculations are essential for stoichiometric conversions needed in quantitative chemical analysis.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions mix, resulting in the formation of an insoluble solid known as a precipitate. In our exercise, \( \mathrm{AgCl} \) is the precipitate formed.
This happens because the ions in the solutions combine to form a compound that is not soluble in water. These reactions are significant for separating and analyzing components in a mixture.
For example, when \( \mathrm{BaCl}_{2}(\mathrm{aq}) \) and \( \mathrm{AgNO}_{3}(\mathrm{aq}) \) react, \( \mathrm{AgCl}(\mathrm{s}) \) is formed as an insoluble solid, appearing as a white precipitate. Observations like these help in understanding the reactivity and compatibility of different ions.

• Importance of such reactions:
  • Helps in purifying mixtures.
  • Used in quantitative analysis to determine the presence of specific ions.

Recognizing and predicting precipitation reactions is fundamental in experimental chemistry and various practical applications.