Question

Balance each chemical equation. \(\mathrm{a} . \mathrm{CO}_{2}(g)+\mathrm{CaSiO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\) b. \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(a q) \longrightarrow \mathrm{Co}_{2} \mathrm{S}_{3}(s)+\mathrm{NH}_{4} \mathrm{NO}_{3}(a q)\) \(\mathrm{c} . \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{CO}(g)\) \(\mathrm{d} . \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g)\)

Short answer

a. \(2\mathrm{CO}_{2}(g)+\mathrm{CaSiO}_{3}(s)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(aq)\) b. \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{3}(aq)+3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(aq) \longrightarrow \mathrm{Co}_{2} \mathrm{S}_{3}(s)+6\mathrm{NH}_{4} \mathrm{NO}_{3}(aq)\) c. \(\mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{C}(s) \longrightarrow 2\mathrm{Cu}(s)+\mathrm{CO}(g)\) d. \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2\mathrm{HCl}(g)\)

Step-by-step solution

Balancing the first chemical equation

To balance the equation \(\mathrm{CO}_{2}(g)+\mathrm{CaSiO}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(aq)\),start by counting the number of atoms of each element on both sides of the equation. Balance the elements one at a time. As Calcium (Ca) appears once on both sides, it is already balanced. Balance Silicon (Si) by having one SiO2 on the right side. Carbon (C) is also balanced with one CO2. Balance the Oxygen (O) and Hydrogen (H) atoms by making sure the number of water molecules and hydrogen carbonates are equal on both sides. In this case, we have 2 hydrogens on the left and 4 hydrogens on the right; adding 2 H20 on the left will balance the Hydrogen. To balance Oxygen, we have 5 Oxygens on the left and 7 on the right; adding one more CO2 on the left will give us 7 Oxygens on each side. The balanced equation is \(2\mathrm{CO}_{2}(g)+\mathrm{CaSiO}_{3}(s)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(aq)\).

Balancing the second chemical equation

Balance the equation \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{3}(aq)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(aq) \longrightarrow \mathrm{Co}_{2} \mathrm{S}_{3}(s)+\mathrm{NH}_{4} \mathrm{NO}_{3}(aq)\)by counting the atoms of each element. Begin by balancing Cobalt (Co) with 2 on both sides. Nitrogen (N) appears as 3 nitrates on the left, while on the right, there are 3 ammonium ions per ammonium nitrate; to balance, place a coefficient of 3 in front of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\). Sulfur (S) is balanced with 3 on the left as part of the sulfide. For Hydrogen (H), we now have 6 from ammonium nitrate and 8 from ammonium sulfide, making a total of 14; place a coefficient of 7 in front of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) to match this number. The balanced equation is \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{3}(aq)+3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(aq) \longrightarrow \mathrm{Co}_{2} \mathrm{S}_{3}(s)+6\mathrm{NH}_{4} \mathrm{NO}_{3}(aq)\).

Balancing the third chemical equation

In the equation \(\mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{CO}(g)\),start by looking at the Copper (Cu) atoms; there are 2 on the left and 1 on the right, thus place a coefficient of 2 in front of Cu on the right. Now Carbon (C) is balanced with one atom on both sides. Oxygen (O) is also balanced with one atom on each side. This gives the balanced equation \(\mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{C}(s) \longrightarrow 2\mathrm{Cu}(s)+\mathrm{CO}(g)\).

Balancing the fourth chemical equation

To balance the equation \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g)\),we see that the hydrogen and chlorine are diatomic molecules. To balance, place a coefficient of 2 in front of HCl, which gives us 2 hydrogen atoms and 2 chlorine atoms on both sides. The balanced equation is \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2\mathrm{HCl}(g)\).

Key concepts

Stoichiometry

At its heart, stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. By studying stoichiometry, we are able to determine how much of a substance is consumed or produced during a reaction.

For example, in the reaction involving carbon dioxide and calcium silicate, understanding stoichiometry helps us calculate the precise amount of water and calcium bicarbonate that will be produced from a known quantity of reactants. The balance in the equation represents the stoichiometric coefficients, which provides us with this crucial molar relationship.

A foundational concept in stoichiometry is the mole, which is a standard unit in chemistry representing a specific number of particles, typically atoms or molecules. We use the mole because dealing with individual atoms is impractical due to their small size and huge numbers. By converting all quantities into moles, the equation can be balanced, and the stoichiometry can then be analyzed.

Chemical Reaction

A chemical reaction is a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Signs of a chemical reaction might include color change, temperature change, gas production, or the formation of a precipitate.

In the provided exercises, we see various types of reactions. Combustion, precipitation, and redox reactions are just a few examples. For instance, the reaction between cobalt nitrate and ammonium sulfide forming cobalt sulfide and ammonium nitrate is a type of double displacement reaction where anions and cations exchange partners.

A fundamental step in understanding a chemical reaction is being able to write and balance the chemical equation that represents it. The balanced chemical equation ensures that the same number of atoms of each element are present on both sides of the equation, in adherence to the law of conservation of mass.

Law of Conservation of Mass

This law states that mass is neither created nor destroyed in a chemical reaction. It was formulated by Antoine Lavoisier in the 18th century, and it's a fundamental principle in chemistry. In practice, this means that the total mass of the reactants must equal the total mass of the products in a chemical reaction.

The balancing of chemical equations, as shown in the provided step-by-step solutions, is a direct application of the law of conservation of mass. Each atom of the reactants must be accounted for among the products. For the simple synthesis reaction where hydrogen gas combines with chlorine gas to form hydrogen chloride, the conservation law dictates that we must have equal numbers of hydrogen and chlorine atoms before and after the reaction, leading to the balanced chemical equation with the correct stoichiometric coefficients.