Question

Balance each of the following equations that describe precipitation reactions. a. \(\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{HNO}_{3}(a q)\) b. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+\) \(\mathrm{HNO}_{3}(a q)\) c. \(\operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\) \(\mathrm{HNO}_{3}(a q)\)

Short answer

The balanced equations are as follows: a. \(2\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+2\mathrm{HNO}_{3}(a q)\) b. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+2\mathrm{HNO}_{3}(a q)\) c. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2\mathrm{HNO}_{3}(a q)\)

Step-by-step solution

(Balancing Equation a)

First, we will balance equation a, which is \(\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{HNO}_{3}(a q)\) For balancing this equation, we can add coefficients in front of the chemical species: \(2\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+2\mathrm{HNO}_{3}(a q)\) Now the equation is balanced, with 2 silver atoms, 2 nitrate ions, 1 sulfate ion, and 2 hydrogen ions on each side.

(Balancing Equation b)

Next, let's balance equation b: \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+ \mathrm{HNO}_{3}(a q)\) In this case, we will need to add a coefficient in front of the nitric acid: \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+ 2\mathrm{HNO}_{3}(a q)\) Now the equation is balanced, with 1 calcium atom, 2 nitrate ions, 1 sulfate ion, and 2 hydrogen ions on each side.

(Balancing Equation c)

Finally, let's balance equation c: \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+ \mathrm{HNO}_{3}(a q)\) Again, we will need to add a coefficient in front of the nitric acid: \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+ 2\mathrm{HNO}_{3}(a q)\) The equation is balanced, with 1 lead atom, 2 nitrate ions, 1 sulfate ion, and 2 hydrogen ions on each side.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry. It ensures that the law of conservation of mass is satisfied. This law states that matter cannot be created or destroyed in a chemical reaction. Instead, each element must have the same number of atoms on both sides of the equation.

When balancing, start by writing the unbalanced equation. Then identify each element and count the atoms of each on the reactant and product sides.

• Use coefficients, which are numbers in front of molecules, to balance the atoms. Adjust these coefficients one element at a time instead of changing subscripts, which can alter the substance's identity.
• It's helpful to leave elements that appear in only one reactant and one product until last. Begin balancing with more complex molecules followed by simpler ones.
• A systematic approach will lead to balanced equations where all atoms match on both sides.

Chemical Equations

Chemical equations represent reactions and use symbols and formulas to depict the reactants (the starting substances) and the products (the substances produced). These equations provide a concise way to illustrate chemical changes.

Substances in chemical equations are often noted with their states:

• (aq) for aqueous, meaning dissolved in water.
• (s) for solid
• (l) for liquid
• (g) for gas

For example, in precipitation reactions, solutions mix to form a solid, or precipitate, which is an important concept in these reactions' balancing and analysis.

Understanding chemical equations involves knowing the symbols for each element, interpreting the meaning of the different states, and recognizing how these symbols come together to portray real-world chemical processes.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced.

The coefficients in a balanced equation are key to stoichiometry. They indicate the ratio in which substances react or are produced. For instance, if the equation reads \(2 \mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}\), we see a 2:1:2 ratio.

• By using these ratios, we can calculate how many moles, mass, or volume of a substance are needed or produced.
• Stoichiometry is used to determine limiting reagents, which are the reactants that run out first, stopping the reaction.
• Additionally, it helps calculate percent yield, revealing the efficiency of a reaction compared to theoretical predictions.

This precise predictive ability is why stoichiometry is vital in laboratory settings and industrial chemical production.