Question

Balance each of the following chemical equations. a. \(\mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{LiCl}(s)\) b. \(\mathrm{Ba}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Ba}_{3} \mathrm{N}_{2}(s)\) c. \(\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) e. \(\operatorname{NiS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NiO}(s)+\mathrm{SO}_{2}(g)\) f. \(\operatorname{CaH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g)\) g. \(\mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) h. \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow \mathrm{B}_{4} \mathrm{C}_{3}(s)+\mathrm{CO}_{2}(g)\)

Short answer

a. \(2\mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{LiCl}(s)\) b. \(3\mathrm{Ba}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Ba}_{3}\mathrm{N}_{2}(s)\) c. \(2\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(2\mathrm{Al}(s)+6\mathrm{HCl}(a q) \rightarrow 2\mathrm{AlCl}_{3}(a q)+3\mathrm{H}_{2}(g)\) e. \(2\operatorname{NiS}(s)+2\mathrm{O}_{2}(g) \rightarrow 2\mathrm{NiO}(s)+2\mathrm{SO}_{2}(g)\) f. \(\operatorname{CaH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g)\) g. \(2\mathrm{H}_{2}(g)+1\mathrm{CO}(g) \rightarrow 1\mathrm{CH}_{3}\mathrm{OH}(l)\) h. \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+6\mathrm{C}(s) \rightarrow 2\mathrm{B}_{4}\mathrm{C}_{3}(s)+3\mathrm{CO}_{2}(g)\)

Step-by-step solution

a. Balancing \(\mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{LiCl}(s)\)

Since there are two chlorine atoms in \(\mathrm{Cl}_{2}\) and only one in \(\mathrm{LiCl}\), we need to balance the chlorine atoms by placing a coefficient of 2 in front of \(\mathrm{LiCl}\). This gives us: \(\mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{LiCl}(s)\) Now, we need to balance the lithium atoms by placing a coefficient of 2 in front of \(\mathrm{Li}\). The balanced equation is: \(2\mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{LiCl}(s)\)

b. Balancing \(\mathrm{Ba}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Ba}_{3}\mathrm{N}_{2}(s)\)

Since there are two nitrogen atoms in \(\mathrm{N}_{2}\) and also two nitrogen atoms in \(\mathrm{Ba}_{3}\mathrm{N}_{2}\), we need to balance the barium atoms by placing a coefficient of 3 in front of \(\mathrm{Ba}\). The balanced equation is: \(3\mathrm{Ba}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Ba}_{3}\mathrm{N}_{2}(s)\)

c. Balancing \(\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

To balance the sodium atoms, we need to place a coefficient of 2 in front of \(\mathrm{NaHCO}_{3}\): \(2\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Since there are now two hydrogen atoms in \(2\mathrm{NaHCO}_{3}\) and two hydrogen atoms in \(\mathrm{H}_{2} \mathrm{O}(g)\), the equation is balanced.

d. Balancing \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\)

We start by balancing the chlorine atoms, which requires placing a coefficient of 3 in front of \(\mathrm{HCl}\): \(\mathrm{Al}(s)+3\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) Now, balance the hydrogen atoms by placing a coefficient of 3/2 in front of \(\mathrm{H}_{2}\). This leads to: \(\mathrm{Al}(s)+3\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\frac{3}{2}\mathrm{H}_{2}(g)\) Since we cannot have a fraction as a coefficient, we can multiply the entire equation by 2 to get whole number coefficients: \(2\mathrm{Al}(s)+6\mathrm{HCl}(a q) \rightarrow 2\mathrm{AlCl}_{3}(a q)+3\mathrm{H}_{2}(g)\)

e. Balancing \(\operatorname{NiS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NiO}(s)+\mathrm{SO}_{2}(g)\)

Start by balancing the nickel atoms with coefficients: \(\operatorname{NiS}(s)+1\mathrm{O}_{2}(g) \rightarrow 1\mathrm{NiO}(s)+1\mathrm{SO}_{2}(g)\) To balance the oxygen atoms, we find the least common multiple of the coefficients, which is 2. Then, we multiply the whole equation by 2: \(2\operatorname{NiS}(s)+2\mathrm{O}_{2}(g) \rightarrow 2\mathrm{NiO}(s)+2\mathrm{SO}_{2}(g)\)

f. Balancing \(\operatorname{CaH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g)\)

The equation is already balanced.

g. Balancing \(\mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{CH}_{3}\mathrm{OH}(l)\)

Place coefficients in front of \(\mathrm{H}_{2}\) and \(\mathrm{CO}\) to balance hydrogen and carbon atoms: \(2\mathrm{H}_{2}(g)+1\mathrm{CO}(g) \rightarrow 1\mathrm{CH}_{3}\mathrm{OH}(l)\)

h. Balancing \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow \mathrm{B}_{4}\mathrm{C}_{3}(s)+\mathrm{CO}_{2}(g)\)

First, balance the boron atoms, which requires a coefficient of 2 for \(\mathrm{B}_{4}\mathrm{C}_{3}\): \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow 2\mathrm{B}_{4}\mathrm{C}_{3}(s)+\mathrm{CO}_{2}(g)\) Next, balance the carbon atoms by placing a coefficient of 6 in front of \(\mathrm{C}\): \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+6\mathrm{C}(s) \rightarrow 2\mathrm{B}_{4}\mathrm{C}_{3}(s)+\mathrm{CO}_{2}(g)\) Finally, balance the oxygen atoms by placing a coefficient of 3 in front of \(\mathrm{CO}_{2}\): \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+6\mathrm{C}(s) \rightarrow 2\mathrm{B}_{4}\mathrm{C}_{3}(s)+3\mathrm{CO}_{2}(g)\)

Key concepts

Understanding Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is a key concept for students to understand as it allows them to calculate how much of a reactant is needed to produce a desired amount of product, or how much product can be formed from a given amount of reactant.

The crucial aspect of stoichiometry is based on the law of conservation of mass, which states that in any chemical reaction, matter is neither created nor destroyed. This law implies that the amount of each element must be the same in the reactants and products. To achieve this balance, you would adjust the coefficients (the numbers in front of the chemical formulas) in a reaction equation to reflect the correct ratio of molecules.

• For example, when balancing the equation for the reaction of lithium (Li) with chlorine gas (Cl₂) to form lithium chloride (LiCl), the stoichiometry shows that two lithium atoms react with one molecule of chlorine gas to produce two units of lithium chloride.
• In more complex reactions, such as when sodium hydrogen carbonate (NaHCO₃) decomposes into sodium carbonate (Na₂CO₃), carbon dioxide gas (CO₂), and water vapor (H₂O), it becomes increasingly important to understand the stoichiometric relationships to balance the equation correctly.

For students, mastering stoichiometry entails practice in recognizing the molar ratios and using them to solve problems involving mass-to-mass conversions, volume-to-volume conversions, and determining limiting reactants in a chemical reaction.

Chemical Reactions

Chemical reactions involve the breaking and forming of bonds between atoms, resulting in the transformation of substances. This transformation is represented by a chemical equation, which depicts the reactants, substances that undergo a chemical change, and the products, new substances formed as a result of the reaction.

Understanding chemical reactions is fundamental for students of chemistry. Each reaction is governed by certain rules and principles, such as the law of conservation of mass and the need to have the same number of each type of atom on both sides of the equation.

Type of Chemical Reactions

Chemical reactions can be categorized into several types:

• Synthesis: Two or more reactants combine to form a single product.
• Decomposition: A single compound breaks down into two or more simpler substances.
• Single replacement: One element replaces another in a compound.
• Double replacement: Two compounds exchange ions or atoms to form two new compounds.
• Combustion: A substance reacts with oxygen to produce heat and light.

For instance, the decomposition of sodium hydrogen carbonate into sodium carbonate, carbon dioxide, and water vapor, as shown in one of the exercises, is an example of a decomposition reaction. By studying the types of chemical reactions and the way they are represented in chemical equations, students can better understand the process of balancing these equations and predicting the outcomes of reactions.

Coefficients in Chemistry

Coefficients in chemistry are numerical values used in chemical equations to indicate the number of molecules or moles of each substance participating in a reaction. By adjusting these coefficients, chemists ensure that the number of atoms for each element is equal on both sides of the equation, thus obeying the law of conservation of mass.

In a balanced chemical equation, the coefficients represent the smallest whole number ratio of reactants and products. These ratios are referred to as stoichiometric coefficients and are essential in stoichiometric calculations, allowing chemists to quantify the amounts of substances consumed and produced.

Significance of Coefficients

• Coefficients indicate the proportions of reactants and products involved in the reaction.
• By maintaining these proportions, the reaction can be completely understood in terms of the quantity of substances used and formed.
• They provide the conversion factor between moles of one substance to moles of another in a balanced reaction.

For example, in the reaction between aluminum (Al) and hydrochloric acid (HCl) to form aluminum chloride (AlCl₃) and hydrogen gas (H₂), the coefficient of 2 in front of the Al in the balanced equation indicates that two moles of aluminum react with six moles of hydrochloric acid to produce two moles of aluminum chloride and three moles of hydrogen gas. Students' understanding of coefficients is critical for accurately interpreting and carrying out reactions in both the classroom and the lab.