Problem 6 - 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: \(\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) (a) Write the balanced equation. (b) What mass of \(\mathrm{AgNO}_{3},\) in grams, is required for complete reaction with \(0.156 \mathrm{g}\) of \(\mathrm{BaCl}_{2} ?\) What mass of AgCl is produced?

Step-by-Step Solution

Answer

To react with 0.156 g of BaCl2, you need 0.2547 g of AgNO3 and 0.2147 g of AgCl is produced.

Step 1: Balance the Chemical Equation

To balance the equation, ensure each type of atom has the same number on both sides. Start by writing: \[ \text{BaCl}_2(\text{aq}) + 2\text{AgNO}_3(\text{aq}) \rightarrow 2\text{AgCl}(\text{s}) + \text{Ba(NO}_3\text{)}_2(\text{aq}) \] Here, the chloride and silver numbers are balanced using 2 moles of AgNO3 and AgCl, while barium and nitrate remain balanced.

Step 2: Calculate Molar Masses

Calculate the molar mass of each compound: - BaCl2: 1(137.33) + 2(35.45) = 208.23 g/mol - AgNO3: 1(107.87) + 1(14.01) + 3(16.00) = 169.87 g/mol - AgCl: 1(107.87) + 1(35.45) = 143.32 g/mol.

Step 3: Determine Moles of BaCl2

Find the number of moles of BaCl2 using its given mass: \[ \text{moles of BaCl}_2 = \frac{0.156 \text{ g}}{208.23 \text{ g/mol}} = 0.000749 \text{ moles} \]

Step 4: Use Stoichiometry for AgNO3

According to the balanced equation, 1 mole of \(\text{BaCl}_2\) reacts with 2 moles of \(\text{AgNO}_3\). Calculate moles of \(\text{AgNO}_3\): \[ \text{moles of AgNO}_3 = 2 \times 0.000749 = 0.001498 \text{ moles} \]

Step 5: Calculate Mass of AgNO3 Required

Convert moles of \(\text{AgNO}_3\) to grams: \[ \text{mass of AgNO}_3 = 0.001498 \text{ moles} \times 169.87 \text{ g/mol} = 0.2547 \text{ g} \]

Step 6: Calculate Mass of AgCl Produced

For every mole of \(\text{BaCl}_2\), 2 moles of \(\text{AgCl}\) are produced. Calculate mass: - Moles of \(\text{AgCl}\): 0.001498 - Mass: \[ \text{mass of AgCl} = 0.001498 \times 143.32 \text{ g/mol} = 0.2147 \text{ g} \]

Key Concepts

Chemical Reactions

Chemical reactions are fundamental to the field of chemistry, encompassing the process by which substances, known as reactants, transform into new substances, called products. During a chemical reaction, molecules rearrange their atoms, leading to changes in their chemical properties. The initial and final substances can differ in physical state, color, and even in temperature. This change occurs through breaking and forming of chemical bonds. A common way to represent a chemical reaction is through a chemical equation that displays the reactants and products with their respective chemical formulas.

• Reactants are substances that start a chemical reaction.
• Products are the new substances formed as a result of the reaction.
• Conditions such as temperature and pressure can influence the rate and outcome of chemical reactions.

Students learning chemistry need to grasp how reactions proceed, their components, and how they are energetically driven.

Balancing Chemical Equations

Balancing chemical equations is a critical skill in chemistry, aiming to demonstrate the conservation of mass principle. This principle states that atoms are neither created nor destroyed in a chemical reaction. Therefore, a balanced equation has the same number of each type of atom on both sides.
Consider the balanced reaction: \[ \text{BaCl}_2(\text{aq}) + 2\text{AgNO}_3(\text{aq}) \rightarrow 2\text{AgCl}(\text{s}) + \text{Ba(NO}_3\text{)}_2(\text{aq})\]Here, you balance:

• Chlorine and silver by using coefficients of 2 for \(\text{AgNO}_3\) and \(\text{AgCl}\).
• Nitrate ions \(\text{NO}_3^-\) are also balanced, ensuring equal numbers on both sides.
• Atoms of barium remain unchanged since \(\text{BaCl}_2\) and \(\text{Ba(NO}_3\text{)}_2\) naturally balance each other.

By adjusting these coefficients, the equation complies with the law of conservation of mass, offering a clear picture of the stoichiometry involved in the reaction.

Molar Mass Calculation

Calculating molar mass is an essential part of quantitative chemistry, enabling students to convert between grams and moles, a common requirement in stoichiometry. Molar mass represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To calculate the molar mass, sum the atomic masses of all the atoms present in a molecule, considering their respective quantities. For example, computing the molar mass for the compounds involved entails the following:

• For \(\text{BaCl}_2\), use atomic weights: Barium (137.33), Chlorine (35.45).
• For \(\text{AgNO}_3\), consider Silver (107.87), Nitrogen (14.01), Oxygen (16.00).
• For \(\text{AgCl}\), sum up Silver (107.87) and Chlorine (35.45).

These calculations provide the masses: \(208.23\,\text{g/mol}\) for \(\text{BaCl}_2\), \(169.87\,\text{g/mol}\) for \(\text{AgNO}_3\), and \(143.32\,\text{g/mol}\) for \(\text{AgCl}\). Knowing these values helps in determining the quantities necessary for reacting substances and predicting the amount of product formed.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where insoluble solids, known as precipitates, form when two aqueous solutions combine. They are an essential aspect of analytical chemistry because they allow scientists to separate and identify components of mixtures. In the reaction provided, when \(\text{BaCl}_2\) and \(\text{AgNO}_3\) react, \(\text{AgCl}\) precipitates as a solid because it is not soluble in water.

• These reactions are usually double displacement reactions, involving an exchange of ions between the reacting species.
• The formed precipitate can be predicted using solubility rules, which guide whether a compound remains dissolved or forms a solid.
• In laboratory settings, precipitates are often filtered and analyzed to quantify the extent of the reaction.

Understanding precipitation reactions enables students to comprehend how ionic compounds interact in various solutions, providing a practical insight into salt and mineral formation.