Problem 47 - Chemistry Matter and Change

Question

When hydrochloric acid solution reacts with lead(II) nitrate solution, lead(II) chloride precipitates and a solution of nitric acid is produced. a. Write the balanced chemical equation for the reaction. b. Interpret the equation in terms of molecules and formula units, moles, and mass.

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between hydrochloric acid and lead(II) nitrate is: \( 2HCl(aq) + Pb(NO_{3})_{2}(aq) \rightarrow PbCl_{2}(s) + 2HNO_{3}(aq) \) Interpretation: - Molecules and formula units: 2 molecules of HCl react with 1 formula unit of Pb(NO₃)₂ to produce 1 formula unit of PbCl₂ and 2 molecules of HNO₃. - Moles: 2 moles of HCl react with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbCl₂ and 2 moles of HNO₃. - Mass: 73 g of HCl react with 331.29 g of Pb(NO₃)₂ to produce 278.11 g of PbCl₂ and 126.02 g of HNO₃.

Step 1: Write the reactants and products

Based on the exercise, the reactants are hydrochloric acid (HCl) and lead(II) nitrate (Pb(NO₃)₂). The products will be lead(II) chloride (PbCl₂) and nitric acid (HNO₃).

Step 2: Balance the chemical equation

Write the unbalanced equation as: HCl + Pb(NO₃)₂ → PbCl₂ + HNO₃ Balance all atoms, other than Oxygen and Hydrogen first, in this case, lead (Pb), and chloride (Cl): HCl + Pb(NO₃)₂ → PbCl₂ + HNO₃ H and Cl: 2HCl + Pb(NO₃)₂ → PbCl₂ + 2HNO₃ Now the equation is balanced: 2HCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2HNO₃(aq)

Step 3: Interpret the equation in terms of molecules and formula units, moles, and mass

Begin by looking at the balanced chemical equation: 2HCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2HNO₃(aq) Molecules and formula units: - Two molecules of hydrochloric acid react with one formula unit of lead(II) nitrate to produce one formula unit of lead(II) chloride and two molecules of nitric acid. Moles: - 2 moles of hydrochloric acid react with 1 mole of lead(II) nitrate to produce 1 mole of lead(II) chloride and 2 moles of nitric acid. Mass: By calculating each compound's molar mass, we can write: (2 × 36.5 g/mol HCl) + (1 × 331.29 g/mol Pb(NO₃)₂) → (1 × 278.11 g/mol PbCl₂) + (2 × 63.01 g/mol HNO₃) Thus, 73 g of HCl reacted with 331.29 g of Pb(NO₃)₂ results in 278.11g of PbCl₂ and 126.02 g of HNO₃.

Key Concepts

Balancing Chemical Equations

At the core of understanding chemical reactions, is learning how to balance chemical equations. It’s not enough to know that substances react; it's essential to know in what ratios they combine and the resulting products.

When you balance a chemical equation, you ensure that the number of each type of atom on the reactant side equals the number of those atoms on the product side. This is because in a chemical reaction, matter is neither created nor destroyed; it simply changes form according to the Law of Conservation of Mass.

Let’s review our example involving hydrochloric acid and lead(II) nitrate. Initially, the equation is written as: HCl + Pb(NO₃)₂ → PbCl₂ + HNO₃. However, this does not accurately reflect the reactants converting to products in a 1-to-1 ratio. To balance the equation, you have to work systematically, typically balancing atoms of metals, then non-metals, and lastly, hydrogen and oxygen.

Here's how we achieve balance in our reaction:

• We note that there are two nitrate (NO₃) groups attached to the lead on the reactant side, and hence we need two nitric acid molecules on the product side.
• We then adjust the chlorine and hydrogen atoms accordingly, which leads to having two hydrochloric acid molecules on the reactant side.

Once balanced, the equation reads: 2HCl + Pb(NO₃)₂ → PbCl₂ + 2HNO₃, meaning for every one lead(II) nitrate molecule, we need two hydrochloric acid molecules to produce one lead(II) chloride molecule and two nitric acid molecules.

Stoichiometry

Stoichiometry is the tool chemists leverage to relate the amounts of reactants to products in a chemical reaction. Stoichiometric calculations allow us to predict the outcomes of reactions, making it a crucial aspect of chemistry studies.

Using our balanced chemical equation, we can understand the proportions of reactants that will react and the amounts of products formed. Stoichiometry relies on the mole concept—wherein one mole equals Avogadro's number (6.022 × 10²³) of entities, whether they are atoms, molecules, ions, or any other chemical units.

In our balanced reaction, the stoichiometry reveals that 2 moles of hydrochloric acid react with 1 mole of lead(II) nitrate to form 1 mole of lead(II) chloride and 2 moles of nitric acid. This becomes incredibly practical when scaling the reaction to actual quantities. For instance, if you have a certain amount of lead(II) nitrate, using stoichiometry, you can determine how much hydrochloric acid you will need and predict how much lead(II) chloride and nitric acid will be produced.

Molar Mass Calculations

With molar mass calculations, we're moving from the world of atoms and moles to the more tangible world of grams and kilograms. Molar mass is the weight of 1 mole of a substance. By knowing the molar masses of each reactant and product, we can convert moles to grams and vice versa, which is helpful for practical laboratory work.

For each substance involved in the reaction, we determine the molar mass by summing the masses of all the atoms in the formula according to the periodic table. For example, hydrochloric acid (HCl) has a molar mass of approximately 36.5 g/mol, derived from adding the mass of hydrogen (about 1 g/mol) to that of chlorine (about 35.5 g/mol).

Practically, if we know the molar masses and the stoichiometry of the reaction, we can calculate not only how many moles of each reactant are needed but also the mass of each reactant required to carry out a reaction at a certain scale, and the mass of each product we expect to produce. In the exercise provided, this knowledge allows us to translate our balanced chemical equation into mass terms for the lab—a fundamental step for any chemist.