Problem 65 - Chemistry Matter and Change

Question

A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate. \begin{equation} \begin{array}{l}{\text { a. Write the balanced chemical equation. }} \\\ {\text { b. Starting with } 0.250 \text { mol of potassium chromate, }} \\\ {\text { determine the mass of lead chromate formed. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

80.8 grams of lead(II) chromate is formed.

Step 1 - Write the Unbalanced Chemical Equation

First, identify the chemical formulas of the reactants and products. Potassium chromate is \( \text{K}_2\text{CrO}_4 \) and lead(II) nitrate is \( \text{Pb(NO}_3\text{)}_2 \). The products are lead(II) chromate \( \text{PbCrO}_4 \) and potassium nitrate \( \text{KNO}_3 \). The unbalanced equation is: \[ \text{K}_2\text{CrO}_4 + \text{Pb(NO}_3\text{)}_2 \rightarrow \text{PbCrO}_4 + \text{KNO}_3 \]

Step 2 - Balance the Chemical Equation

To balance the equation, ensure that there are equal numbers of each type of atom on both sides. Balancing the equation gives: \[ \text{K}_2\text{CrO}_4 + \text{Pb(NO}_3\text{)}_2 \rightarrow \text{PbCrO}_4 + 2 \text{KNO}_3 \]

Step 3 - Calculate Moles of Lead(II) Chromate

Given 0.250 mol of \( \text{K}_2\text{CrO}_4 \), use the stoichiometry from the balanced equation. Since the molar ratio of \( \text{K}_2\text{CrO}_4 \) to \( \text{PbCrO}_4 \) is 1:1, you also have 0.250 mol of \( \text{PbCrO}_4 \).

Step 4 - Calculate the Molar Mass of Lead(II) Chromate

Calculate the molar mass of \( \text{PbCrO}_4 \) using the atomic masses: \( \text{Pb} = 207.2 \), \( \text{Cr} = 52.0 \), \( \text{O} = 16.0 \). Thus, the molar mass is \( 207.2 + 52.0 + 4 \times 16.0 = 323.2 \, \text{g/mol} \).

Step 5 - Determine the Mass of Lead(II) Chromate Formed

Multiply the moles of \( \text{PbCrO}_4 \) by its molar mass to find the mass: \( 0.250 \, \text{mol} \times 323.2 \, \text{g/mol} = 80.8 \, \text{g} \).

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry. It ensures that the law of conservation of mass is upheld, meaning the amount of each element is consistent on both sides of a chemical equation. In our example, we started with an unbalanced equation featuring the reactants potassium chromate (\( \text{K}_2\text{CrO}_4 \)) and lead(II) nitrate (\( \text{Pb(NO}_3\text{)}_2 \)) leading to the products lead(II) chromate (\( \text{PbCrO}_4 \)) and potassium nitrate (\( \text{KNO}_3 \)).

Initially, writing the chemical formulas correctly is essential. Once that's done, balance the equation by adjusting coefficients. Here, the balanced equation becomes \[ \text{K}_2\text{CrO}_4 + \text{Pb(NO}_3\text{)}_2 \rightarrow \text{PbCrO}_4 + 2 \text{KNO}_3 \] as we adjust chromium and nitrate ions to appear equally on both sides.

• Always check the number of atoms for each element.
• Use coefficients to balance different species, not changing the subscripts in a compound formula.

By balancing the equation, you set a foundation for quantitative chemistry analyses like stoichiometry.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. In this reaction, it helps us calculate exactly how much lead(II) chromate we can expect to form from a given amount of potassium chromate.

The key is in the molar ratios derived from the balanced equation. Since our chemical equation is \( 1:1 \) with respect to \( \text{K}_2\text{CrO}_4 \) and \( \text{PbCrO}_4 \):

• Start with 0.250 mol of \( \text{K}_2\text{CrO}_4 \)
• The same number of moles of lead(II) chromate, \( \text{PbCrO}_4 \), will be produced because of the one-to-one ratio that stoichiometry provides.

This straightforward relationship makes stoichiometry a powerful tool in predictive reactions and resource management in laboratories.

Molar Mass

Molar mass is the mass of a given substance divided by its amount of substance, measured in moles. It is critical when converting between moles and grams, central to solving many chemistry problems. For lead(II) chromate, \( \text{PbCrO}_4 \), determining its molar mass means summing its atoms' masses based on the periodic table.

Here's how we compute it:

• Lead (Pb) contributes \( 207.2 \) g/mol.
• Chromium (Cr) contributes \( 52.0 \) g/mol.
• Each oxygen (O) atom adds \( 16.0 \) g/mol, and there are four: \( 4 \times 16.0 = 64.0 \) g/mol.

Thus, adding these numbers you get \( 323.2 \) g/mol for lead(II) chromate.

Finally, to find the mass of \( \text{PbCrO}_4 \) produced, multiply its molar mass by the moles present. Thus, \( 0.250 \text{ mol} \times 323.2 \text{ g/mol} = 80.8 \text{ g} \). Understanding molar mass is essential for translating chemical formulae into measurable realities.