Question

Balance each chemical equation. a. \(\mathrm{Na}_{2} \mathrm{S}(a q)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{NaNO}_{3}(a q)+\mathrm{CuS}(s)\) b. \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\) c. \(\operatorname{HCl}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) \(\mathrm{d} . \mathrm{FeS}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{S}(g)\)

Short answer

The balanced chemical equations are: (a) \(2\mathrm{Na}_{2} \mathrm{S}(aq)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq) \longrightarrow 2\mathrm{NaNO}_{3}(aq)+\mathrm{CuS}(s)\) (b) \(3\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4\mathrm{NH}_{3}(g)+3\mathrm{N}_{2}(g)\) (c) \(4\mathrm{HCl}(aq)+\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+2\mathrm{Cl}_{2}(g)\) (d) \(\mathrm{FeS}(s)+2\mathrm{HCl}(aq) \longrightarrow \mathrm{FeCl}_{2}(aq)+\mathrm{H}_{2} \mathrm{S}(g)\).

Step-by-step solution

Balance Equation (a)

To balance the chemical equation \(\mathrm{Na}_{2} \mathrm{S}(aq)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq) \longrightarrow \mathrm{NaNO}_{3}(aq)+\mathrm{CuS}(s)\), start by balancing the most complex molecule. In this case, that's \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}. There's one \mathrm{Cu} atom on both sides, so the copper is already balanced. There are two \mathrm{NO}_{3} groups on the left, so you need two \mathrm{NaNO}_{3} on the right for balance. This results in two sodium atoms on the right, meaning you also need two \mathrm{Na}_{2} \mathrm{S} on the left side to balance the sodium. The balanced equation becomes \(2\mathrm{Na}_{2} \mathrm{S}(aq)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq) \longrightarrow 2\mathrm{NaNO}_{3}(aq)+\mathrm{CuS}(s)\).

Balance Equation (b)

In the decomposition reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\), start by balancing the nitrogen atoms. There are 2 nitrogen atoms in \(\mathrm{N}_{2} \mathrm{H}_{4}\) and two in \(\mathrm{N}_{2}\), so the nitrogens are balanced. There are 4 hydrogen atoms on the left but only 3 on the right within the ammonia molecules, so you need to multiply \mathrm{NH}_{3} by 2 to balance the hydrogens. This gives you a total of 6 hydrogens on the right, so you need to have 3 \(\mathrm{N}_{2} \mathrm{H}_{4}\) on the left to get 6 hydrogens there as well. The balanced equation is \(3\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4\mathrm{NH}_{3}(g)+3\mathrm{N}_{2}(g)\).

Balance Equation (c)

The equation \(\mathrm{HCl}(aq)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) involves balancing oxygen, hydrogen, and chlorine. First, balance the oxygen by making sure there are 2 oxygen atoms on each side, thus requiring 2 \(\mathrm{H}_{2}\mathrm{O}\) molecules on the right. Now there are 4 hydrogens on the right, which means you need 4 \mathrm{HCl} molecules on the left. This results in 2 chlorine atoms on the left, requiring 1 molecule of \mathrm{Cl}_{2} on the right for balance. The balanced equation is \(4\mathrm{HCl}(aq)+\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)+2\mathrm{Cl}_{2}(g)\).

Balance Equation (d)

For the reaction \(\mathrm{FeS}(s)+\mathrm{HCl}(aq) \longrightarrow \mathrm{FeCl}_{2}(aq)+\mathrm{H}_{2} \mathrm{S}(g)\), you start by checking the balance of atoms on both sides of the equation. Both iron (Fe) and sulfur (S) are already balanced with one atom each on both sides. However, there are two chlorine (Cl) atoms in \(\mathrm{FeCl}_{2}\) on the right side, so you need 2 \mathrm{HCl} on the left to balance the chlorine, which then gives you 2 hydrogen (H) atoms on the left side. Since \(\mathrm{H}_{2} \mathrm{S}\) contains 2 H atoms, the hydrogen is balanced as well. The balanced equation is \(\mathrm{FeS}(s)+2\mathrm{HCl}(aq) \longrightarrow \mathrm{FeCl}_{2}(aq)+\mathrm{H}_{2} \mathrm{S}(g)\).

Key concepts

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is fundamental to balance chemical equations accurately. It ensures the reaction adheres to the Law of Conservation of Mass, where mass can neither be created nor destroyed. Stoichiometry comes from Greek words that mean 'measuring elements'; it allows chemists to predict the amounts of substances consumed and produced in a given reaction. For instance, when balancing the decomposition reaction in Exercise b, stoichiometry guides us to adjust coefficients to ensure there are equal numbers of each type of atom on both sides of the equation. Remember, it's not just about balancing atoms, but also about reflecting the actual quantities that react and form products in predictable proportions.

Decomposition Reaction

A decomposition reaction involves a single compound breaking down into two or more products, typically requiring energy in the form of heat, light, or electricity. In the textbook example b, hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), decomposes into ammonia, \(\mathrm{NH}_{3}\), and nitrogen gas,\(\mathrm{N}_{2}\). This is a classic decomposition reaction where the complex molecule splits apart. When balancing decomposition reactions, it is crucial to consider all atoms in the molecule, and sometimes the products dictate the need for coefficients to ensure mass balance, as seen in the three \(\mathrm{N}_{2} \mathrm{H}_{4}\) molecules producing four \(\mathrm{NH}_{3}\) molecules and three \(\mathrm{N}_{2}\) molecules.

Chemical Equation Notation

Proper chemical equation notation is essential for communicating the specifics of a chemical reaction. This includes not only the chemical formulas of reactants and products but also their physical states, indicated by abbreviations in parentheses: \(s\) for solids, \(l\) for liquids, \(g\) for gases, and \(aq\) for aqueous solutions. For example, in the response to exercise a, \(\mathrm{Na}_{2} \mathrm{S}(aq)\) refers to the aqueous solution of sodium sulfide, while \(\mathrm{CuS}(s)\) signifies the solid form of copper sulfide. By using these annotations, chemists provide a clear picture of the reaction conditions.

Law of Conservation of Mass

The law of conservation of mass is a guiding principle in chemical reactions and mandates that all atoms present in the reactants must be accounted for in the products. This is why chemists must balance chemical equations, ensuring the same number of each type of atom appears on both sides of the reaction. For instance, in exercise c, the chlorine atoms were balanced by doubling \(\mathrm{HCl}\) and \(\mathrm{Cl}_{2}\) to keep the total number of chlorine atoms consistent, reflecting that matter has neither been gained nor lost, just reorganized. Understanding this law is critical, as it reveals the fundamental constancy of matter during physical transformations and chemical reactions.