Question

Balance each of the following equations: a. \(\mathrm{Ca}(s)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{CaBr}_{2}(s)\) b. \(\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)\) c. \(\mathrm{Sb}_{2} \mathrm{~S}_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{SbCl}_{3}(s)+\mathrm{H}_{2} \mathrm{~S}(g)\) d. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}(g)\)

Short answer

a. Already balanced. b. \(\mathrm{P}_{4}(s)+5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s)\) c. \(\mathrm{Sb}_{2}\mathrm{~S}_{3}(s)+6\mathrm{HCl}(aq) \longrightarrow 2 \mathrm{SbCl}_{3}(s)+3\mathrm{H}_{2}\mathrm{~S}(g)\) d. \(\mathrm{Fe}_{2}\mathrm{O}_{3}(s)+3\mathrm{C}(s)\longrightarrow 2\mathrm{Fe}(s)+3\mathrm{CO}(g)\)

Step-by-step solution

Understand the Problem

The goal is to balance the chemical equations by making sure the number of atoms of each element on the reactant side equals the number of atoms on the product side.

Equation a: \(\mathrm{Ca}(s)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{CaBr}_{2}(s)\)

1. List the number of atoms of each element in the reactants and products. Reactants: 1 Ca, 2 Br Products: 1 Ca, 2 Br 2. The number of atoms on both sides are already equal. Thus, the equation is balanced. Balanced Equation: \(\mathrm{Ca}(s)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{CaBr}_{2}(s)\)

Equation b: \(\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s)\)

1. List the number of atoms of each element in the reactants and products. Reactants: 4 P, 2 O Products: 4 P, 10 O 2. Balance the oxygen atoms by putting a coefficient of 5 in front of \(\mathrm{O}_{2}\). Balanced Equation: \(\mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) \)

Equation c: \(\mathrm{Sb}_{2}\mathrm{~S}_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{SbCl}_{3}(s) + \mathrm{H}_{2}\mathrm{~S}(g)\)

1. List the number of atoms of each element in the reactants and products. Reactants: 2 Sb, 3 S, 1 H, 1 Cl Products: 1 Sb, 1 Cl, 2 H, 1 S 2. Balance the Sb atoms by putting a coefficient of 2 in front of \(\mathrm{SbCl}_{3}\). Now we have: 2 \(\mathrm{Sb}\), 3 \(\mathrm{S}\), 6 \(\mathrm{H}\), 6 \(\mathrm{Cl}\) 3. Balance H and Cl atoms by putting a coefficient of 6 in front of \(\mathrm{HCl}\). Balanced Equation: \(\mathrm{Sb}_{2}\mathrm{~S}_{3}(s)+6\mathrm{HCl}(a q) \longrightarrow 2 \mathrm{SbCl}_{3}(s)+3\mathrm{H}_{2}\mathrm{~S}(g)\)

Equation d: \(\mathrm{Fe}_{2}\mathrm{O}_{3}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}(g)\)

1. List the number of atoms of each element in the reactants and products. Reactants: 2 Fe, 3 O, 1 C Products: 1 Fe, 1 O, 1 C 2. Balance the Fe atoms by putting a coefficient of 2 in front of \(\mathrm{Fe}\). Now we have: Reactants: 2 Fe, 3 O, 1 C Products: 2 Fe, 1 O, 1 C 3. Balance the oxygen atoms by putting a coefficient of 3 in front of \(\mathrm{CO}\). Now we have: Reactants: 2 Fe, 3 O, 1 C Products: 2 Fe, 3 O, 3 C 4. Balance the carbon atoms by putting a coefficient of 3 in front of \(\mathrm{C}\): Balanced Equation: \(\mathrm{Fe}_{2}\mathrm{O}_{3}(s)+3\mathrm{C}(s)\longrightarrow 2\mathrm{Fe}(s)+3\mathrm{CO}(g) \)

Key concepts

stoichiometry

Stoichiometry is the heart of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Essentially, stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction. When balancing equations, stoichiometry ensures that the proportion of atoms of each element is the same on both sides of the reaction.
For instance, in our example: \[ \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) \longrightarrow 2\mathrm{Fe}(s) + 3\mathrm{CO}(g) \] stoichiometry helps us determine exactly how much carbon is needed to fully react with the iron(III) oxide, ensuring no excess reactants are left. This can be visualized as a recipe where changing the amount of one ingredient requires adjusting others to maintain the desired outcome.

law of conservation of mass

The Law of Conservation of Mass is a fundamental principle stating that mass is neither created nor destroyed in a chemical reaction. This implies the total mass of reactants equates to the total mass of products.
When balancing a chemical equation, this law underscores the need to ensure the same number of atoms for each element on both sides of the equation. Take the balance of \[ \mathrm{Ca}(s) + \mathrm{Br}_{2}(l) \longrightarrow \mathrm{CaBr}_{2}(s) \] as an example. Here, we have one calcium atom and two bromine atoms on the reactant side, matched by one calcium atom and two bromine atoms on the product side, illustrating the law perfectly.
This concept is critical in chemical reactions and helps in maintaining consistency across chemical processes, ensuring no atom is lost in the transition from reactants to products.

chemical reactions

Chemical reactions are processes where reactants transform into products through the breaking and forming of chemical bonds. These reactions are represented through chemical equations indicating the species involved and their quantities.
Reactions can be diverse, ranging from simple synthesis reactions like \[ \mathrm{Ca}(s) + \mathrm{Br}_{2}(l) \longrightarrow \mathrm{CaBr}_{2}(s) \] to more complex ones like \[ \mathrm{Fe}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) \longrightarrow 2\mathrm{Fe}(s) + 3\mathrm{CO}(g) \]. Each type of reaction follows specific patterns and laws, such as the conservation of mass and energy.
Understanding how to balance these reactions is essential as it ensures the chemical equation accurately represents the actual process, reflecting the proportions of reactants and products involved.

reactants and products

In every chemical reaction, substances known as reactants undergo transformations to form new substances called products. Reactants are the starting materials, while products are the end results of the reaction.
For example, in the equation \[ \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \], phosphorus (\mathrm{P}_{4}) and oxygen (\mathrm{O}_{2}) are the reactants, while diphosphorus pentoxide (\mathrm{P}_{4} \mathrm{O}_{10}) is the product.
Recognizing the role of each species in a reaction is crucial for correctly balancing the equation and understanding the stoichiometric relationships that define the reaction's outcome. This comprehension aids in predicting reaction behavior and the quantities of products formed from given reactants.