Question

Balance each of the following chemical equations by inspection: (a) \(\mathrm{HClO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{~s}) \rightarrow \mathrm{Ba}\left(\mathrm{ClO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g) \rightarrow \mathrm{CoS}(s)+\mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{Fe}(\mathrm{s})+\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Cd}(s)\) (d) \(\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}_{2}(g)\) (e) \(\operatorname{Sn}(s)+\mathrm{P}(s) \rightarrow \operatorname{Sn}_{3} \mathrm{P}_{2}(s)\)

Short answer

(a) 2 HClO4 + Ba(OH)2 -> Ba(ClO4)2 + 2 H2O; (b) Co(NO3)2 + H2S -> CoS + 2 HNO3; (c) 2 Fe + 3 Cd(NO3)2 -> 2 Fe(NO3)3 + 3 Cd; (d) Fe2(CO3)3 -> Fe2O3 + 3 CO2; (e) 3 Sn + 2 P -> Sn3P2.

Step-by-step solution

Analyzing Equation (a)

The equation given is \( \mathrm{HClO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{~s}) \rightarrow \mathrm{Ba}\left(\mathrm{ClO}_{4}\right)_{2}(s)+\mathrm{H}_{2}\mathrm{O}(l) \). Start by balancing the elements one at a time. Here, we need to balance perchlorate (\( \text{ClO}_4^- \)), barium (\( \text{Ba}^{2+} \)), and hydroxide (\( \text{OH}^- \)). We have 2 \( \text{ClO}_4^- \) on the right, which means we need 2 \( \text{HClO}_4 \) on the left. For \( \text{Ba}(\text{OH})_2 \), everything is already balanced.

Balanced Equation (a)

The balanced equation is \( 2 \mathrm{HClO}_{4}(a q) + \mathrm{Ba}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Ba}\left(\mathrm{ClO}_{4}\right)_{2}(s) + 2 \mathrm{H}_{2}\mathrm{O}(l) \).

Analyzing Equation (b)

The equation is \( \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q) + \mathrm{H}_{2}\mathrm{~S}(g) \rightarrow \mathrm{CoS}(s) + \mathrm{HNO}_{3}(a q) \). First, balance cobalt (\( \text{Co} \)) and sulfur (\( \text{S} \)) using \( \text{CoS} \) and \( \text{H}_2 \text{S} \). Next, balance the nitrate (\( \text{NO}_3^- \)); you need 2 \( \text{HNO}_3 \) to match with the 2 nitrate groups on Co(NO3)2.

Balanced Equation (b)

The balanced equation is \( \mathrm{Co}\left( \mathrm{NO}_{3} \right)_{2} (a q) + \mathrm{H}_{2}\mathrm{S} (g) \rightarrow \mathrm{CoS} (s) + 2 \mathrm{HNO}_{3} (aq) \).

Analyzing Equation (c)

The equation is \( \mathrm{Fe}(\mathrm{s}) + \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q) + \mathrm{Cd}(s) \). Balance iron (\( \text{Fe} \)) and cadmium (\( \text{Cd} \)) first. Notice \( \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} \) contains 3 nitrate groups, so you need 3 \( \text{Cd}\left(\mathrm{NO}_{3}\right)_{2} \) and therefore 3 cadmium atoms on the right. Also, match iron using 2 atoms.

Balanced Equation (c)

The balanced equation is \( 2 \mathrm{Fe} (s) + 3 \mathrm{Cd}\left( \mathrm{NO}_{3} \right)_{2} (a q) \rightarrow 2 \mathrm{Fe}\left( \mathrm{NO}_{3} \right)_{3} (aq) + 3 \mathrm{Cd}(s) \).

Analyzing Equation (d)

The unbalanced equation is \( \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + \mathrm{CO}_{2}(g) \). Start by balancing iron (\( \text{Fe}_2 \)), which is already balanced. Next, balance the carbonate ion (\( \text{CO}_3^{2-} \)). Since there are 3 carbonate ions, this produces 3 \( \text{CO}_2 \).

Balanced Equation (d)

The balanced equation is \( \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 3 \mathrm{CO}_{2}(g) \).

Analyzing Equation (e)

The equation \( \operatorname{Sn}(s) + \mathrm{P}(s) \rightarrow \operatorname{Sn}_{3}\mathrm{P}_{2}(s) \) needs to have both elements balanced. We need 3 tin (\( \text{Sn} \)) and 2 phosphorus (\( \text{P} \)) on both sides.

Balanced Equation (e)

The balanced equation is \( 3 \operatorname{Sn}(s) + 2 \mathrm{P}(s) \rightarrow \operatorname{Sn}_{3}\mathrm{P}_{2}(s) \).

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. This transformation involves the breaking and forming of chemical bonds, leading to changes at the molecular level. A classic example is

1. An acid-base reaction where an acid reacts with a base to form water and a salt.
2. Another is a redox reaction where electron transfer occurs between two species.

In the provided exercises, we are dealing with various types of reactions: including acid-base, redox, and synthesis reactions. Balancing these reactions is a key skill, allowing you to predict the outcomes of the reactions and understand the stoichiometry, which refers to the quantitative relationships involved. Always ensure each atom's count is equal on both sides of the equation, reflecting the Law of Conservation of Mass.

Stoichiometry

Stoichiometry is a section of chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and the concept of the mole, which is a fundamental unit in chemistry used to express amounts of a chemical substance. When balancing chemical reactions, stoichiometry helps to quantify materials:

• First, identify the number of moles of each reactant and product.
• Then, use coefficients to balance the equation, which shows the ratio in which substances react.

For example, in the reaction \[ 2 \mathrm{HClO}_{4}(a q) + \mathrm{Ba}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Ba}\left(\mathrm{ClO}_{4}\right)_{2}(s) + 2 \mathrm{H}_{2}\mathrm{O}(l) \],this equation tells us that 2 moles of perchloric acid react with 1 mole of barium hydroxide to produce 1 mole of barium perchlorate and 2 moles of water. Stoichiometry allows chemists to scale this reaction up or down depending on the amount of reactants available.

Inorganic Chemistry

Inorganic chemistry is the study of the properties and behavior of inorganic compounds, which include minerals, metals, and nonmetals. Inorganic chemistry often involves:

• Studying the synthesis and behavior of inorganic and organometallic compounds.
• Understanding reactions such as co-ordination, precipitation, and redox reactions.

In the context of the exercises provided, inorganic chemistry principles are crucial. For example, understanding the reaction:\[\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 3 \mathrm{CO}_{2}(g)\]is essential for learning about decomposition reactions, which are common in inorganic chemistry. These reactions lead to the breakdown of compounds into simpler substances, often involving sharing or transferring of electrons. Mastery of these concepts allows better understanding of material creation and transformation on a molecular level, providing insights into industrial processes and new material development.