Question

Balance the following equations: (a) \(\mathrm{CF}_{4}(l)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CBr}_{4}(l)+\mathrm{F}_{2}(g)\) (b) \(\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{MnO}_{2}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) (d) \(\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

The balanced equations are: (a) \(\mathrm{CF}_{4}(l)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CBr}_{4}(l)+2\mathrm{F}_{2}(g)\) (b) \(\mathrm{Cu}(s)+8\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{NO}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{MnO}_{2}(s)+4\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+2\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) (d) \(3\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-step solution

Identify elements

First, let's identify the elements in the equation: - Carbon (C) - Fluorine (F) - Bromine (Br)

Count atoms of each element on both sides

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 C, 4 F, and 2 Br - On the right side: 1 C, 4 Br, and 2 F

Balance the equation

Since there are 4 F on the left side and 2 F on the right side, we need to add a coefficient of 2 in front of \(\mathrm{F}_{2}\) on the right side of the equation. Now our equation looks like: \(\mathrm{CF}_{4}(l)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CBr}_{4}(l)+2\mathrm{F}_{2}(g)\) At this point, the equation is balanced. (b) Balancing the equation $\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$:

Identify elements

First, let's identify the elements in the equation: - Copper (Cu) - Hydrogen (H) - Nitrogen (N) - Oxygen (O)

Count atoms of each element on both sides

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 Cu, 1 H, 1 N, and 3 O - On the right side: 1 Cu, 2 H, 5 N, and 9 O

Balance the equation

Since there are 1 N and 3 O on the left side, and 5 N and 9 O on the right side, we will need to add a coefficient of 8 in front of \(\mathrm{HNO}_{3}\) on the left side of the equation: $\mathrm{Cu}(s)+8\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ Now, we need to add a coefficient of 2 in front of \(\mathrm{NO}_{2}\) and a coefficient of 4 in front of \(\mathrm{H}_{2} \mathrm{O}\) on the right side of the equation to balance the atoms: $\mathrm{Cu}(s)+8\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{NO}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(l)$ At this point, the equation is balanced. (c) Balancing the equation $\mathrm{MnO}_{2}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$:

Identify elements

First, let's identify the elements in the equation: - Manganese (Mn) - Oxygen (O) - Hydrogen (H) - Chlorine (Cl)

Count atoms of each element on both sides

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 Mn, 2 O, 1 H, and 1 Cl - On the right side: 1 Mn, 1 O, 2 H, and 3 Cl

Balance the equation

Since there are 1 H and 1 Cl on the left side, and 2 H and 3 Cl on the right side, we will add a coefficient of 4 in front of \(\mathrm{HCl}\) on the left side of the equation: $\mathrm{MnO}_{2}(s)+4\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$ And after that, place a coefficient of 2 in front of \(\mathrm{H}_{2} \mathrm{O}\) and a coefficient of 1 in front of \(\mathrm{Cl}_{2}(g)\) on the right side of the equation: $\mathrm{MnO}_{2}(s)+4\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+2\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$ At this point, the equation is balanced. (d) Balancing the equation $\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$:

Identify elements

First, let's identify the elements in the equation: - Potassium (K) - Hydrogen (H) - Oxygen (O) - Phosphorus (P)

Count atoms of each element on both sides

Count the number of atoms for each element on both sides of the equation: - On the left side: 1 K, 4 H, 1 P, and 4 O - On the right side: 3 K, 2 H, 1 P, and 4 O

Balance the equation

Since there are 1 K and 4 H on the left side, and 3 K and 2 H on the right side, we will add a coefficient of 3 in front of \(\mathrm{KOH}\) on the left side of the equation: $3\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ After that, place a coefficient of 3 in front of \(\mathrm{H}_{2} \mathrm{O}\) on the right side of the equation: $3\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)$ At this point, the equation is balanced.

Key concepts

Stoichiometry

Stoichiometry is a vital concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows chemists to predict how much of each substance is needed or produced in a given reaction. This is essential for understanding how reactions occur and ensuring that resources are used efficiently.

In stoichiometry, balanced chemical equations are crucial. They show the exact proportions of molecules involved, achieved by using coefficients in front of chemical formulas to balance the number of each type of atom on both reactant and product sides. Without balancing, reactions cannot follow the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction.

• When balancing, one should ensure that the same number of each type of atom appears on both sides.
• Consider a balanced equation like \( ext{2H}_2 + ext{O}_2 \rightarrow ext{2H}_2 ext{O}\) where 4 hydrogen and 2 oxygen atoms are present on each side.
• Stoichiometric calculations help in discovering the limiting reactant, which determines the amount of product formed.

Understanding stoichiometry is key to solving chemical equations, optimizing reactions, and is foundational for advanced topics like reaction yield calculations.

Chemical Reactions

Chemical reactions are transformative processes where reactants are converted into products. They are fundamental to chemistry and involve changes in the arrangement of atoms that result in new substances. During a chemical reaction, bonds between atoms are broken and new bonds are formed, often accompanied by energy changes.

Chemical reactions can be categorized into several types such as synthesis, decomposition, single replacement, and double replacement. In balancing chemical reactions, it’s crucial to accurately represent the reactants and products and understand the type of reaction occurring.

• In synthesis reactions, multiple reactants combine to form a single product, for instance, \( ext{A} + ext{B} \rightarrow ext{AB}\).
• Decomposition reactions involve a single compound breaking down into two or more simpler substances, like \( ext{AB} \rightarrow ext{A} + ext{B}\).
• In single replacement reactions, one element replaces another in a compound, while double replacement involves an exchange of elements between two compounds.

Recognizing the type of chemical reaction assists in balancing equations and predicting reaction behavior. Mastery of chemical reactions is vital for fields ranging from biochemistry to industrial processes.

Conservation of Mass

The law of conservation of mass is a foundational principle in chemistry. It states that the mass of reactants in a chemical reaction must equal the mass of the products. This principle is essential for balancing chemical equations and ensuring that no atoms are lost or created in the process.

During a chemical reaction, although the form and appearance of substances change, the total mass remains constant. This is because chemical reactions only reorganize the existing atoms into new compounds.

• A balanced chemical equation reflects the conservation of mass by having an equal number of each type of atom on both sides of the equation.
• Balancing involves adjusting coefficients - not the subscripts in chemical formulas, as that would change the actual substances involved.
• For example, balancing \(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\) shows that two nitrogen atoms and six hydrogen atoms are conserved across both sides.

The conservation of mass is not only pivotal in chemistry but also underpins many principles in physics and engineering. Understanding this concept is key to all quantitative chemistry calculations.