Question

Balance each of the following chemical equations by inspection: (a) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\operatorname{NaBr}(a q) \rightarrow \mathrm{Hg}_{2} \mathrm{Br}_{2}(s)+\mathrm{NaNO}_{3}(a q)\) (c) \(\mathrm{Mg}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mg}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) (d) \(\mathrm{LiNO}_{3}(s) \rightarrow \mathrm{LiNO}_{2}(s)+\mathrm{O}_{2}(g)\) (e) \(\mathrm{Pb}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{PbO}(s)\)

Short answer

(a) Already balanced. (b) Balance with 2 NaBr and 2 NaNO₃. (c) Add 2 HC₂H₃O₂. (d) Use 2 LiNO₃ and 2 LiNO₂. (e) Balance with 2 Pb and 2 PbO.

Step-by-step solution

Balance Equation (a)

For the reaction \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\), note that the number of each atom on both sides of the equation is already balanced. Thus, no coefficients are needed. The balanced equation is: \[ \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \] .

Balance Equation (b)

For \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NaBr}(a q) \rightarrow \mathrm{Hg}_{2} \mathrm{Br}_{2}(s)+\mathrm{NaNO}_{3}(a q)\), we start by balancing the bromine atoms. There are 2 bromine atoms needed for \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\), so we place a coefficient of 2 in front of \(\mathrm{NaBr}\). Next, balance the sodium by placing a coefficient of 2 in front of \(\mathrm{NaNO}_{3}\). The balanced equation is: \[ \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{NaBr}(a q) \rightarrow \mathrm{Hg}_{2} \mathrm{Br}_{2}(s)+2\mathrm{NaNO}_{3}(a q) \] .

Balance Equation (c)

For \(\mathrm{Mg}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mg}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2}(g)\), first balance magnesium. Add a coefficient of 2 to \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) because 2 acetate ions are needed to react with one magnesium atom. The balanced equation is: \[ \mathrm{Mg}(s)+2\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mg}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2}(g) \] .

Balance Equation (d)

For \(\mathrm{LiNO}_{3}(s) \rightarrow \mathrm{LiNO}_{2}(s)+\mathrm{O}_{2}(g)\), oxygen needs to be balanced. One molecule of \(\mathrm{O}_{2}\) provides 2 oxygen atoms, while the reactants and \(\mathrm{LiNO}_{2}\) consume only one. Place the coefficient 2 in front of both \(\mathrm{LiNO}_{3}\) and \(\mathrm{LiNO}_{2}\). The balanced equation is: \[ 2\mathrm{LiNO}_{3}(s) \rightarrow 2\mathrm{LiNO}_{2}(s)+\mathrm{O}_{2}(g) \] .

Balance Equation (e)

For \(\mathrm{Pb}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{PbO}(s)\), since each \(\mathrm{O}_{2}\) molecule contains two oxygen atoms, add a coefficient of 2 in front of \(\mathrm{PbO}\) to balance with one \(\mathrm{O}_{2}\) molecule. This adjustment requires adding a coefficient of 2 in front of \(\mathrm{Pb}\) to balance lead atoms. The balanced equation is: \[ 2\mathrm{Pb}(s)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{PbO}(s) \] .

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that allows us to determine the quantities of substances consumed and produced in a chemical reaction. It is essentially the bookkeeping of chemical equations where the coefficients in the balanced equation represent the molar ratios of the reactants and products.

For example, if a balanced chemical equation tells us that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water, stoichiometry helps us calculate exactly how much of each substance is involved and how much is produced.

• Stoichiometry provides the framework to understand the proportions of chemicals that participate in reactions.
• It ensures that chemical reactions follow the law of conservation of mass.
• Understanding stoichiometry is crucial for tasks like calculating yields and determining limiting reactants.

By applying stoichiometry, chemists can predict the amounts of products formed or the quantities of reactants required.

Reaction Coefficients

Reaction coefficients are the numbers placed before compounds in a chemical equation to balance the equation and make sure that the same number of each type of atom appears on both sides.

These coefficients allow for the representation of the stoichiometric relationships between reactants and products. For instance, in the balanced equation for the reaction between hydrogen gas and oxygen gas to form water, the coefficients are 2 for hydrogen, 1 for oxygen, and 2 for water: \[ 2 ext{H}_2 + ext{O}_2 ightarrow 2 ext{H}_2 ext{O} \]

Reaction coefficients tell us:

• How many molecules (or moles) of a substance take part in the reaction.
• The ratio in which different substances react and are produced.
• Are essential for calculating reagent quantities and reaction yields.

Thus, understanding and correctly applying reaction coefficients is essential for balancing chemical equations and performing calculations in stoichiometry.

Chemical Reactions

Chemical reactions are processes where substances (reactants) transform into new substances (products). These transformations involve the breaking and forming of chemical bonds. A basic example is the combustion of methane:

\[ ext{CH}_4 + 2 ext{O}_2 ightarrow ext{CO}_2 + 2 ext{H}_2 ext{O} \]

In this reaction, methane burns in the presence of oxygen, producing carbon dioxide and water. Key components of chemical reactions include:

• Reactants: The starting substances that undergo change.
• Products: The new substances formed.
• Energy Change: Reactions typically involve energy exchange, either absorbing or releasing heat.

Chemical reactions can be classified based on the nature of the changes, such as synthesis, decomposition, single replacement, or double replacement. Understanding chemical reactions requires recognizing these transformations and balancing the resulting equations to adhere to the laws of chemistry.

Conservation of Mass

The conservation of mass is a core principle in chemistry stating that mass cannot be created or destroyed in a chemical reaction. Instead, it is conserved, meaning the total mass of the reactants equals the total mass of the products. This principle is ensured through the proper balancing of chemical equations.

For example, in a simple balanced reaction like: \[ ext{N}_2 + 3 ext{H}_2 ightarrow 2 ext{NH}_3 \]

The number of nitrogen and hydrogen atoms is the same on both sides of the equation, illustrating that the mass remains constant throughout the reaction. Here’s what to remember about conservation of mass:

• It is the reason we balance chemical equations.
• Ensures that atoms are neither lost nor gained.
• Helps in predicting the quantities of substances in a reaction.

Emphasizing the conservation of mass is essential for understanding and predicting the outcomes of chemical reactions reliably.