Question

Balance each of the following chemical equations. a. \(\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{KCl}(a q)\) b. \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{FeO}(s)+\mathrm{H}_{2}(g)\) c. \(\mathrm{NaOH}(a q)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{NaClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\operatorname{Mg}(s)+\operatorname{Mn}_{2} \mathrm{O}_{3}(s) \rightarrow \mathrm{MgO}(s)+\operatorname{Mn}(s)\) e. \(\mathrm{KOH}(s)+\mathrm{KH}_{2} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) f. \(\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HNO}_{3}(a q)\) g. \(\mathrm{BaO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{O}_{2}(g)\) h. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

The balanced chemical equations are: a. \( \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+2\mathrm{KCl}(a q) \) b. \( \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{FeO}(s)+\mathrm{H}_{2}(g) \) c. \( \mathrm{NaOH}(a q)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{NaClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \) d. \( \mathrm{Mg}(s)+\mathrm{Mn}_{2} \mathrm{O}_{3}(s) \rightarrow \mathrm{MgO}(s)+2\mathrm{Mn}(s) \) e. \( 2\mathrm{KOH}(s)+\mathrm{KH}_{2} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3\mathrm{H}_{2} \mathrm{O}(l) \) f. \( 3\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{HNO}_{3}(a q) \) g. \( \mathrm{BaO}_{2}(s)+2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{O}_{2}(g) \) h. \( 4\mathrm{NH}_{3}(g)+5\mathrm{O}_{2}(g) \rightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2} \mathrm{O}(l) \)

Step-by-step solution

a. Balancing the first equation

In this equation, the reactants are \(\mathrm{K}_{2} \mathrm{SO}_{4}\) and \(\mathrm{BaCl}_{2}\), while the products are \(\mathrm{BaSO}_{4}\) and \(\mathrm{KCl}\). First, balance the barium (Ba) atoms and the potassium (K) atoms, then balance the remaining elements: sulfur (S), oxygen (O), and chlorine (Cl) atoms. \( \underline{1}\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\underline{1}\mathrm{BaCl}_{2}(a q) \rightarrow \underline{1}\mathrm{BaSO}_{4}(s)+\underline{2}\mathrm{KCl}(a q) \)

b. Balancing the second equation

In this equation, the reactants are \(\mathrm{Fe}\) and \(\mathrm{H}_{2} \mathrm{O}\), while the products are \(\mathrm{FeO}\) and \(\mathrm{H}_{2}\). Begin by balancing the iron (Fe) atoms and the hydrogen (H) atoms, then balance the oxygen (O) atoms. \( \underline{1}\mathrm{Fe}(s)+\underline{1}\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \underline{1}\mathrm{FeO}(s)+\underline{1}\mathrm{H}_{2}(g) \)

c. Balancing the third equation

In this equation, the reactants are \(\mathrm{NaOH}\) and \(\mathrm{HClO}_{4}\), and the products are \(\mathrm{NaClO}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\). Start by balancing the sodium (Na), hydrogen (H), chlorine (Cl), and oxygen (O) atoms. \( \underline{1}\mathrm{NaOH}(a q)+\underline{1}\mathrm{HClO}_{4}(a q) \rightarrow \underline{1}\mathrm{NaClO}_{4}(a q)+\underline{1}\mathrm{H}_{2} \mathrm{O}(l) \)

d. Balancing the fourth equation

In this equation, the reactants are \(\mathrm{Mg}\) and \(\mathrm{Mn}_{2} \mathrm{O}_{3}\), and the products are \(\mathrm{MgO}\) and \(\mathrm{Mn}\). Begin by balancing the magnesium (Mg), manganese (Mn), and oxygen (O) atoms. \( \underline{1}\mathrm{Mg}(s)+\underline{1}\mathrm{Mn}_{2} \mathrm{O}_{3}(s) \rightarrow \underline{1}\mathrm{MgO}(s)+\underline{2}\mathrm{Mn}(s) \)

e. Balancing the fifth equation

In this equation, the reactants are \(\mathrm{KOH}\) and \(\mathrm{KH}_{2} \mathrm{PO}_{4}\), and the products are \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\). Start by balancing the potassium (K), hydrogen (H), phosphorus (P), and oxygen (O) atoms. \( \underline{2}\mathrm{KOH}(s)+\underline{1}\mathrm{KH}_{2} \mathrm{PO}_{4}(a q) \rightarrow \underline{1}\mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\underline{3}\mathrm{H}_{2} \mathrm{O}(l) \)

f. Balancing the sixth equation

In this equation, the reactants are \(\mathrm{NO}_{2}\), \(\mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{O}_{2}\), while the products are \(\mathrm{HNO}_{3}\). Begin by balancing the nitrogen (N), hydrogen (H), and oxygen (O) atoms. \( \underline{3}\mathrm{NO}_{2}(g)+\underline{1}\mathrm{H}_{2} \mathrm{O}(l)+\underline{1}\mathrm{O}_{2}(g) \rightarrow \underline{2}\mathrm{HNO}_{3}(a q) \)

g. Balancing the seventh equation

In this equation, the reactants are \(\mathrm{BaO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), and the products are \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(\mathrm{O}_{2}\). Start by balancing the barium (Ba), hydrogen (H), and oxygen (O) atoms. \( \underline{1}\mathrm{BaO}_{2}(s)+\underline{2}\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \underline{1}\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\underline{1}\mathrm{O}_{2}(g) \)

h. Balancing the eighth equation

In this equation, the reactants are \(\mathrm{NH}_{3}\) and \(\mathrm{O}_{2}\), while the products are \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\). Begin by balancing the nitrogen (N), hydrogen (H), and oxygen (O) atoms. \( \underline{4}\mathrm{NH}_{3}(g)+\underline{5}\mathrm{O}_{2}(g) \rightarrow \underline{4}\mathrm{NO}(g)+\underline{6}\mathrm{H}_{2} \mathrm{O}(l) \)

Key concepts

Stoichiometry

Stoichiometry can be seen as the 'recipe' of chemistry. It's a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In basic terms, it tells you the amounts of substances that react and are produced in a chemical reaction.

When balancing chemical equations, like in the provided exercise, stoichiometry is the tool that ensures the number of atoms for each element is conserved. This conservation reflects the law of conservation of mass, which states that matter cannot be created or destroyed. Stoichiometry provides the coefficients (the numbers placed before the chemical formulas) that balance the equation. These coefficients represent the ratio of molecules that react with each other, and they're crucial for calculations like determining how much product can be formed from a certain amount of reactant, or how much reactant is needed to create a specified amount of product.

To successfully use stoichiometry, one must first achieve a balanced chemical equation, which is the foundation for all subsequent stoichiometric calculations.

Chemical Reactions

Chemical reactions are processes in which substances, the reactants, undergo chemical changes to form new substances, the products. These reactions are represented by chemical equations. Each reactant and product in a chemical equation is represented by its chemical formula, showing the elements and the number of atoms of each element that makes up the compound.

In the exercise, several types of chemical reactions are shown, such as double displacement reactions (as seen in part a of the exercise), combination reactions, and decomposition reactions. Balancing these chemical equations is vital to accurately represent what occurs during the chemical reaction. It's important to note that during a chemical reaction, the bonds between atoms break and new bonds form, which can release or absorb energy, but the atoms themselves don't change; they simply rearrange. The steps provided in the solution ensure that the same number of each type of atom appears on both sides of the arrow, indicating that the atoms have been conserved in the reaction.

Chemical Formula

A chemical formula provides important information about a chemical compound, such as its composition and the ratio of different atoms within the compound. For instance, in the formula \( \mathrm{H}_2\mathrm{O} \), we see that water is made of two hydrogen atoms bonded to one oxygen atom. The numbers in a chemical formula (like the '2' in \( \mathrm{H}_2\mathrm{O} \)) are called subscripts, and they show how many atoms of each element are present in one molecule of the compound.

Understanding chemical formulas is essential for balancing chemical equations because they allow us to count the atoms of each element. When we balance equations, we’re making sure that the number of atoms for each element is the same on both sides of the equation. For instance, in the formula \( \mathrm{KH}_2\mathrm{PO}_4 \) from exercise e, there is one potassium (K), two hydrogens (H), one phosphorus (P), and four oxygen (O) atoms. During the balancing process, these numbers are taken into account to ensure matter is conserved. Without the correct usage of chemical formulas, the stoichiometry of the reaction would be impossible to ascertain.