Question

Balance the following equations: (a) \(\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{HPO}_{3}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) (b) \(\mathrm{Au}_{2} \mathrm{~S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Au}(s)+\mathrm{H}_{2} \mathrm{~S}(g)\) (c) \(\mathrm{Ba}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(a q) \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{NH}_{3}(g)\) (d) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\)

Short answer

The balanced equations are: (a) $\mathrm{HClO}_{4}(aq) + \mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(aq) + 2\mathrm{Cl}_{2}\mathrm{O}_{7}(l)$ (b) $\mathrm{Au}_{2}\mathrm{S}_{3}(s) + 3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s) + 3\mathrm{H}_{2}\mathrm{S}(g)$ (c) $\mathrm{Ba}_{3}\mathrm{N}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(aq) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(aq) + 2\mathrm{NH}_{3}(g)$ (d) $\mathrm{Na}_{2}\mathrm{CO}_{3}(aq) + 2\mathrm{HCl}(aq) \longrightarrow 2\mathrm{NaCl}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)$

Step-by-step solution

Since there is 1 Hydrogen atom on the left side, we need to place a coefficient of 4 before HPO3 on the right side of the equation: $\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(a q)+\mathrm{Cl}_{2}\mathrm{O}_{7}(l)$ #\tag_title# Step 2: Balance the number of Chlorine (Cl) atoms

Now we have 4 chlorine atoms on the left side and 2 on the right side. To balance the number of Cl atoms, we need to place a coefficient of 2 before \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) on the right side of the equation: $\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(a q)+2\mathrm{Cl}_{2}\mathrm{O}_{7}(l)$ #\tag_title# Step 3: Check the O atoms

At this point, we have 16 Oxygen atoms on both sides of the equation, so this equation is now balanced. (b) Balancing the equation: #\tag_title# Step 1: Balance the number of Au atoms

Since there are 2 Au atoms on the left side, we need to place a coefficient of 2 before \(\mathrm{Au}\) on the right side of the equation: $\mathrm{Au}_{2}\mathrm{S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s)+\mathrm{H}_{2}\mathrm{S}(g)$ #\tag_title# Step 2: Balance the number of S atoms

Now we have 3 S atoms on the left side and 1 on the right side. To balance the number of S atoms, we need to place a coefficient of 3 before \(\mathrm{H}_{2}\mathrm{S}\) on the right side of the equation: $\mathrm{Au}_{2}\mathrm{S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s)+3\mathrm{H}_{2}\mathrm{S}(g)$ #\tag_title# Step 3: Balance the number of H atoms

Now we have 2 H atoms on the left side and 6 on the right side. To balance the number of H atoms, we need to place a coefficient of 3 before \(\mathrm{H}_{2}\) on the left side of the equation: $ \mathrm{Au}_{2}\mathrm{S}_{3}(s)+3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s)+3\mathrm{H}_{2}\mathrm{S}(g)$ (c) Balancing the equation: #\tag_title# Step 1: Balance the number of Ba atoms

Since there are 3 Ba atoms on the left side, we need to place a coefficient of 3 before \(\mathrm{Ba}(\mathrm{OH})_{2}\) on the right side of the equation: $\mathrm{Ba}_{3}\mathrm{N}_{2}(s)+\mathrm{H}_{2}\mathrm{O}(a q) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{NH}_{3}(g)$ #\tag_title# Step 2: Balance the number of N atoms

Now we have 2 N atoms on the left side and 1 on the right side. To balance the number of N atoms, we need to place a coefficient of 2 before \(\mathrm{NH}_{3}\) on the right side of the equation: $\mathrm{Ba}_{3}\mathrm{N}_{2}(s)+\mathrm{H}_{2}\mathrm{O}(a q) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(a q)+2\mathrm{NH}_{3}(g)$ #\tag_title# Step 3: Balance the number of H atoms

Now we have 2 H atoms on the left side and 12 on the right side. To balance the number of H atoms, we need to place a coefficient of 6 before \(\mathrm{H}_{2}\mathrm{O}\) on the left side of the equation: $\mathrm{Ba}_{3}\mathrm{N}_{2}(s)+6\mathrm{H}_{2}\mathrm{O}(a q) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(a q)+2\mathrm{NH}_{3}(g)$ (d) Balancing the equation: #\tag_title# Step 1: Balance the number of Na atoms

Since there are 2 Na atoms on the left side, we need to place a coefficient of 2 before \(\mathrm{NaCl}\) on the right side of the equation: $\mathrm{Na}_{2}\mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow 2\mathrm{NaCl}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)$ #\tag_title# Step 2: Balance the number of Cl atoms

Now we have 1 Cl atom on the left side and 2 on the right side. To balance the number of Cl atoms, we need to place a coefficient of 2 before \(\mathrm{HCl}\) on the left side of the equation: $\mathrm{Na}_{2}\mathrm{CO}_{3}(a q)+2\mathrm{HCl}(a q) \longrightarrow 2\mathrm{NaCl}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)$ #\tag_title# Step 3: Balance the number of H atoms

Now we have 4 H atoms on the left side and 2 on the right side. To balance the number of H atoms, we need to place a coefficient of 2 before \(\mathrm{H}_{2}\mathrm{O}\) on the right side of the equation: $\mathrm{Na}_{2}\mathrm{CO}_{3}(a q)+2\mathrm{HCl}(a q) \longrightarrow 2\mathrm{NaCl}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)$. All the equations are now balanced.

Key concepts

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It captures the transformation of reactants into products during a reaction. In chemical equations, reactants are on the left side, while products are on the right. Each element or compound's chemical formula represents different parts of the equation.

Key aspects to consider when working with chemical equations include ensuring correct chemical formulas for the reactants and products and using appropriate symbols like "+" to separate multiple reactants or products and an arrow (\( \rightarrow \)) to indicate the direction of the reaction. For example, in the equation \(\mathrm{HClO}_{4} + \mathrm{P}_{4} \mathrm{O}_{10} \rightarrow \mathrm{HPO}_{3} + \mathrm{Cl}_{2} \mathrm{O}_{7}\), the reactants \(\mathrm{HClO}_{4}\) and \(\mathrm{P}_{4} \mathrm{O}_{10}\) form the products \(\mathrm{HPO}_{3}\) and \(\mathrm{Cl}_{2} \mathrm{O}_{7}\).

This representation allows chemists to visualize and analyze the changes occurring during chemical reactions, aiding in understanding complex processes.

Stoichiometry

Stoichiometry is a branch of chemistry focused on the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is essential for predicting the quantities of materials consumed and produced in reactions.

Using coefficients in a balanced chemical equation, stoichiometry allows you to determine the ratios of molecules involved. For instance, in a reaction involving \(\mathrm{HClO}_{4}\) reacting with \(\mathrm{P}_{4} \mathrm{O}_{10}\), the coefficients tell you how many molecules of each substance react and are produced. This proportional reasoning is crucial when conducting experiments and scaling up reactions in industrial processes.

Calculating moles, masses, and volume relationships are common stoichiometric analyses, allowing chemists to perform precise and accurate reactions.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. During a reaction, reactant molecules interact to form product molecules, resulting in observable changes.

Every chemical reaction can be classified into different types like synthesis, decomposition, combustion, single-displacement, and double-displacement. In reactions, like the given example where \(\mathrm{Au}_{2} \mathrm{~S}_{3}\) reacts with \(\mathrm{H}_{2}\) to form \(\mathrm{Au}\) and \(\mathrm{H}_{2} \mathrm{~S}\), bonds between gold and sulfur break, allowing new bonds between hydrogen and sulfur to form.

These reactions can release or absorb energy. Understanding chemical reactions allows chemists to harness these processes in creating new materials and decomposing existing ones, broadening the scope of what is chemically achievable.

Atom Balancing

Atom balancing is a fundamental step in writing a balanced chemical equation, ensuring that the number of atoms for each element is equal on both sides of the equation. This process is essential due to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

To balance an equation, follow these steps:

• Write the unbalanced equation.
• Count the atoms of each element on both sides.
• Add coefficients to balance the number of atoms for each element.
• Verify that all atoms balance and that the equation is simplified.

For example, in balancing \(\mathrm{Ba}_{3} \mathrm{~N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) reacting to form \(\mathrm{Ba(OH)}_{2}\) and \(\mathrm{NH}_{3}\), coefficients are adjusted to achieve equal numbers of each atom on both sides. Balancing ensures a proper depiction of the reaction, allowing for accurate stoichiometric calculations.