Problem 24 - Chemistry for Allied Health

Question

Balance the following equations. a. \(\mathrm{Zn}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{HBr} \rightarrow \mathrm{CaBr}_{2}+\mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) e. \(\mathrm{NH}_{3}+\mathrm{CuO} \rightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The balanced equations are: a) 1:2:1:1, b) 6:1:2, c) 1:2:1:2, d) 2:13:8:10, e) 2:3:3:1:3.

Step 1: Balancing Reaction a

The unbalanced equation is \(\mathrm{Zn}(s) + \mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl}_{2}(aq) + \mathrm{H}_{2}(g)\). To balance the chlorine and hydrogen atoms, we need two \(\mathrm{HCl}\) molecules to react with each zinc atom:Balanced equation: \[\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl}_{2}(aq) + \mathrm{H}_{2}(g)\]

Step 2: Balancing Reaction b

The unbalanced equation is \(\mathrm{Li}(s) + \mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\). There are three lithium atoms needed to balance with one \(\mathrm{N}_{2}\) molecule:Balanced equation: \[ 6\mathrm{Li}(s) + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{Li}_{3}\mathrm{N}(s) \]

Step 3: Balancing Reaction c

The unbalanced equation is \(\mathrm{Ca(OH)}_{2} + \mathrm{HBr} \rightarrow \mathrm{CaBr}_{2} + \mathrm{H}_{2} \mathrm{O}\). To balance the bromine and hydrogen atoms, we need two \(\mathrm{HBr}\) for each \(\mathrm{Ca(OH)}_{2}\):Balanced equation: \[ \mathrm{Ca(OH)}_{2} + 2\mathrm{HBr} \rightarrow \mathrm{CaBr}_{2} + 2\mathrm{H}_{2} \mathrm{O} \]

Step 4: Balancing Reaction d

The unbalanced equation is \(\mathrm{C}_{4}\mathrm{H}_{10} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\). Balance carbon by placing 4 \(\mathrm{CO}_{2}\), hydrogen by placing 5 \(\mathrm{H}_{2}\mathrm{O}\), and finally oxygen:Balanced equation: \[ 2\mathrm{C}_{4}\mathrm{H}_{10} + 13\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + 10\mathrm{H}_{2} \mathrm{O} \]

Step 5: Balancing Reaction e

The unbalanced equation is \(\mathrm{NH}_{3} + \mathrm{CuO} \rightarrow \mathrm{Cu} + \mathrm{N}_{2} + \mathrm{H}_{2} \mathrm{O}\). Balance nitrogen with an \(\mathrm{N}_{2}\), then adjust copper. Place a 3 on \(\mathrm{CuO}\) and \(\mathrm{Cu}\) and adjust ammonia:Balanced equation: \[ 2\mathrm{NH}_{3} + 3\mathrm{CuO} \rightarrow 3\mathrm{Cu} + \mathrm{N}_{2} + 3\mathrm{H}_{2} \mathrm{O} \]

Key Concepts

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships or ratios between the reactants and products in a chemical reaction. It can be thought of as the cookbook for chemistry that guides us on how much of each ingredient is needed to complete a chemical recipe.
In stoichiometry, we use coefficients in a balanced chemical equation to represent these ratios. These coefficients tell us how many molecules or moles of each substance are involved in the reaction. This is crucial when it comes to balancing chemical equations, as it ensures that the number of atoms for each element is the same on both sides of the equation.
For example, in equation (a) \[\mathrm{Zn}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl}_{2}(aq) + \mathrm{H}_{2}(g)\]we see that one mole of zinc reacts with two moles of hydrochloric acid to produce one mole of zinc chloride and one mole of hydrogen gas. The coefficients provide a clear map of the proportions of reactants and products.

Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry stating that matter cannot be created or destroyed in an isolated system. This principle is vital in chemical equations.
During a chemical reaction, even though the substances involved may change form, the total mass of the reactants equals the total mass of the products. This is why balancing equations is so crucial. By ensuring the same number of each type of atom appears on both sides of the equation, we uphold this law.
Take for instance, reaction (d):\[2\mathrm{C}_{4}\mathrm{H}_{10} + 13\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + 10\mathrm{H}_{2}\mathrm{O}\]Here, the mass of all carbon, hydrogen, and oxygen atoms present in the reactants equals the mass of those atoms in the products, adhering to the law of conservation of mass.

Molecular Equations

A molecular equation is a balanced chemical equation where the reactants and products are written as if they were molecules, using the full chemical formulas. Unlike ionic or net ionic equations that show ions separately, molecular equations provide a broad view of the substances involved in the reaction.
This type of equation is essential to understanding stoichiometry and conservation of mass. For example, when writing a molecular equation, all substances are represented form-wise with coefficients included. They provide a straightforward blueprint of the reaction mechanism.
For example, in equation (c):\[\mathrm{Ca(OH)}_{2} + 2\mathrm{HBr} \rightarrow \mathrm{CaBr}_{2} + 2\mathrm{H}_{2} \mathrm{O}\]Every molecule is represented in its entirety, giving us a clear list of what substances start and end the reaction.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They depict the substances involved—from reactants (starting materials) to products (end substances). Understanding chemical equations is pivotal to grasping how chemical reactions occur and are balanced.
Each chemical equation consists of the chemical formulas of the reactants on the left, the products on the right, and an arrow indicating the direction of the reaction. Balancing these equations ensures that the same number of atoms for each element appear on both sides of the reaction, maintaining the conservation of mass.
For instance, in reaction (b):\[6\mathrm{Li}(s) + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{Li}_{3}\mathrm{N}(s)\]The equation shows that lithium and nitrogen react to form lithium nitride. The coefficients are adjusted to balance the elements involved, demonstrating the equation's adherence to stoichiometric principles and the conservation of mass.