Question

Balance the following equations: a. \(\mathrm{Cr}(s)+\mathrm{S}_{\mathrm{s}}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)\) b. \(\mathrm{NaHCO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{KClO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. \(\operatorname{Eu}(s)+\mathrm{HF}(g) \rightarrow \operatorname{EuF}_{3}(s)+\mathrm{H}_{2}(g)\)

Short answer

The balanced equations are: a. \(2\ \text{Cr(s)} + 3\ \text{S(s)} \rightarrow \text{Cr}_2\text{S}_3\text{(s)}\) b. \(2\ \text{NaHCO}_3\text{(s)} \stackrel{\text { Heat }}{\longrightarrow} \text{Na}_2\text{CO}_3\text{(s)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(g)}\) c. \(2\ \text{KClO}_3\text{(s)} \stackrel{\text { Heat }}{\longrightarrow} 2\ \text{KCl(s)} + 3\ \text{O}_2\text{(g)}\) d. \(\text{Eu(s)} + 2\ \text{HF(g)} \rightarrow \text{EuF}_3\text{(s)} + \text{H}_2\text{(g)}\)

Step-by-step solution

a. Balancing the first equation: Cr(s) + S(s) → Cr₂S₃(s)

First, examine the number of chromium (Cr) and sulfur (S) atoms on both the reactant and product side. We have: \(1\ Cr \rightarrow 2\ Cr\), \(1\ S \rightarrow 3\ S\). To balance the chromium atoms, multiply the Cr on the reactant side by 2: \(2\ Cr(s) + S(s) \rightarrow Cr_{2}S_{3}(s)\). Now, to balance the sulfur atoms, multiply S on the reactant side by 3: \(2\ Cr(s) + 3\ S(s) \rightarrow Cr_{2}S_{3}(s)\). The balanced equation is: \(2\ \text{Cr(s)} + 3\ \text{S(s)} \rightarrow \text{Cr}_2\text{S}_3\text{(s)}\).

b. Balancing the second equation: NaHCO₃(s) → Heat → Na₂CO₃(s) + CO₂(g) + H₂O(g)

Examine the number of atoms of each element on both the reactant and product side. We have: \(1\ Na \rightarrow 2\ Na\), \(1\ H \rightarrow 2\ H\), \(1\ C \rightarrow 3\ C\), \(3\ O \rightarrow 7\ O\). To balance the sodium (Na) atoms, multiply the NaHCO₃ on the reactant side by 2: \(2\ NaHCO_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} Na_{2}CO_{3}(s) + CO_{2}(g) + H_{2}O(g)\). Now, the number of carbon (C) atoms and hydrogen (H) atoms are also balanced, but the number of oxygen (O) atoms is still unbalanced. We have: \(6\ O \rightarrow 7\ O\). In this case, since all coefficients are already at their lowest values, the equation is already balanced. The balanced equation is: \(2\ \text{NaHCO}_3\text{(s)} \stackrel{\text { Heat }}{\longrightarrow} \text{Na}_2\text{CO}_3\text{(s)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(g)}\).

c. Balancing the third equation: KClO₃(s) → Heat → KCl(s) + O₂(g)

Examine the number of atoms of each element on both the reactant and product side. We have: \(1\ K \rightarrow 1\ K\), \(1\ Cl \rightarrow 1\ Cl\), \(3\ O \rightarrow 2\ O\). To balance the oxygen (O) atoms, multiply the O₂ on the product side by \(\frac{3}{2}\): \(KClO_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} KCl(s) + \frac{3}{2} O_{2}(g)\). However, since we want to have whole number coefficients for each species, multiply the whole equation by 2: \(2\ KClO_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} 2\ KCl(s) + 3\ O_{2}(g)\). The balanced equation is: \(2\ \text{KClO}_3\text{(s)} \stackrel{\text { Heat }}{\longrightarrow} 2\ \text{KCl(s)} + 3\ \text{O}_2\text{(g)}\).

d. Balancing the fourth equation: Eu(s) + HF(g) → EuF₃(s) + H₂(g)

Examine the number of atoms of each element on both the reactant and product side. We have: \(1\ Eu \rightarrow 1\ Eu\), \(1\ H \rightarrow 2\ H\), \(1\ F \rightarrow 3\ F\). To balance the hydrogen (H) atoms, multiply the HF on the reactant side by 2: \(Eu(s) + 2\ HF(g) \rightarrow EuF_{3}(s) + H_{2}(g)\). Now, the hydrogen atoms are balanced, and we need to balance the fluorine (F) atoms. The coefficients are already at their lowest values, and the equation is already balanced. The balanced equation is: \(\text{Eu(s)} + 2\ \text{HF(g)} \rightarrow \text{EuF}_3\text{(s)} + \text{H}_2\text{(g)}\).

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that relates to calculating the quantities of reactants and products involved in a chemical reaction.

It is akin to a recipe in cooking—just as a recipe specifies the exact amount of each ingredient needed to make a dish, stoichiometry provides the precise measure of each element or compound necessary to ensure a complete reaction.

When balancing chemical equations, stoichiometry is crucial as it ensures the law of conservation of mass is upheld. This law states that matter cannot be created or destroyed; therefore, the quantity of each element must be the same before and after a reaction takes place.

Using stoichiometry, we can determine how to scale amounts for a reaction to proceed, for example, how much of a reactant is needed to produce a desired amount of product, or conversely, how much product we can expect from a given quantity of reactant.

Stoichiometry rests on the mole concept, the units used in chemical equations, and the mole ratios, often derived from a balanced chemical equation, to predict the outcomes of chemical reactions.

Chemical Reaction

A chemical reaction is a process that involves the rearrangement of the molecular or ionic structure of substances. During a reaction, the reactants convert into products, usually with observable changes such as heat production, color change, gas emission, or precipitate formation.

For a reaction to be accurately depicted, a chemical equation is used, which shows the reactants on the left side of an arrow and the products on the right. Balancing this equation is essential because it reflects the conservation of mass, as no atoms are lost or gained in the reaction, only reorganized.

To ensure the equation holds this law of conservation of mass, stoichiometry comes into play, letting us adjust coefficients—simple whole numbers placed before the chemical formulas—to equalize the number of atoms of each element on both sides of the equation.

Understanding chemical reactions helps chemists control conditions to achieve desired products efficiently and predict the side products that could be generated.

Mole Concept

The mole concept is a fundamental principle in chemistry that links the microscopic world of atoms and molecules to the macroscopic world we can measure. One mole consists of exactly 6.022 x 10²³ particles, which could be atoms, molecules, electrons, or other entities, and is known as Avogadro's number.

A vital aspect of the mole concept is that a mole of any element or compound has a mass in grams that is equivalent to its relative atomic or molecular mass. For example, carbon has an atomic mass of ~12, meaning one mole of carbon atoms weighs roughly 12 grams.

In stoichiometric calculations, the mole serves as a bridge, allowing chemists to convert between mass, number of particles, and volume of gases (at standard temperature and pressure) in chemical equations. It is particularly useful because chemical reactions often involve quantities of substances too large to consider at an atomic scale, and counting individual atoms or molecules is impractical.