Problem 102 - Chemistry An Atoms First Approach

Question

Balance the following equations: a. \(\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{S}_{3}(s)\) 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. \(\mathrm{Eu}(s)+\mathrm{HF}(g) \rightarrow \mathrm{EuF}_{3}(s)+\mathrm{H}_{2}(g)\)

Step-by-Step Solution

Answer

The balanced equations are: a. \(6 \operatorname{Cr}(s) + 3S_{8}(s) \rightarrow 8Cr_{2}S_{3}(s)\) b. \(2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \longrightarrow \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(2 \mathrm{Eu}(s) + 6\mathrm{HF}(g) \rightarrow 2\mathrm{EuF}_{3}(s) + 3\mathrm{H}_{2}(g)\)

Step 1: Balancing Chromium

Begin by balancing the Cr atoms. We have 1 Cr atom in the reactants and 2 Cr atoms in the products. To balance the equation, we can put a coefficient of 2 in front of Cr on the reactants side: \(2 \: \operatorname{Cr}(s) + S_8(s) \rightarrow Cr_2S_3(s)\).

Step 2: Balancing Sulfur

Now, let's balance the S atoms. Currently, we have 8 S atoms in the reactants and 3 S atoms in the products. To balance the sulfur atoms, we can put a coefficient of \(\frac{8}{3}\) in front of Cr_2S_3, which gives us: \(2 \operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \frac{8}{3}\mathrm{Cr}_{2} \mathrm{S}_{3}(s)\). However, using fractional coefficients is not ideal, so we can fix that by multiplying the entire equation by 3. This results in the balanced equation: \(6 \operatorname{Cr}(s) + 3\mathrm{S}_{8}(s) \rightarrow 8\mathrm{Cr}_{2} \mathrm{S}_{3}(s)\). #b. Balancing NaHCO_3(s) -> Na_2CO_3(s) + CO_2(g) + H_2O(g)#

Step 1: Balancing Sodium

There is 1 Na atom in the reactant NaHCO_3 and 2 Na atoms in the product Na_2CO_3. We can balance the sodium atoms by putting a coefficient of 2 in front of NaHCO_3: \(2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\).

Step 2: Balancing Carbon

With the current coefficients, there are 2 C atoms on the reactants side and 2 C atoms on the products side, so the carbon atoms are already balanced.

Step 3: Balancing Hydrogen and Oxygen

Finally, we need to balance the H and O atoms. There are 2 H atoms in the reactants and 2 H atoms in the products, so the hydrogen atoms are balanced. There are 6 O atoms in the reactants and 6 O atoms in the products, so the oxygen atoms are also balanced. The balanced equation is: \(2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). #c. Balancing KClO_3(s) -> KCl(s) + O_2(g)#

Step 1: Balancing Potassium and Chlorine

The K and Cl atoms are already balanced, as there is 1 K and 1 Cl atom on each side of the equation.

Step 2: Balancing Oxygen

There are 3 O atoms in the reactant KClO_3 and 2 O atoms in the product O_2. To balance the oxygen atoms, we can use a coefficient of 2 for KClO_3 and a coefficient of 3 for O_2: \(2 \mathrm{KClO}_{3}(s) \longrightarrow \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\). #d. Balancing Eu(s) + HF(g) -> EuF_3(s) + H_2(g)#

Step 1: Balancing Europium

The Eu atoms are already balanced, as there is 1 Eu atom on each side of the equation.

Step 2: Balancing Fluorine

There is 1 F atom in the reactant HF and 3 F atoms in the product EuF_3. To balance the fluorine atoms, we can use a coefficient of 3 for HF: \(\mathrm{Eu}(s)+3 \mathrm{HF}(g) \longrightarrow \mathrm{EuF}_{3}(s)+\mathrm{H}_{2}(g)\).

Step 3: Balancing Hydrogen

Now, there are 3 H atoms in the reactants and 2 H atoms in the products. To achieve balance, we can use a coefficient of \(\frac{3}{2}\) for H_2. However, fractional coefficients are not ideal, so we can multiply the entire equation by 2 to eliminate the fraction: \(2 \mathrm{Eu}(s) + 6\mathrm{HF}(g) \rightarrow 2\mathrm{EuF}_{3}(s) + 3\mathrm{H}_{2}(g)\).

Key Concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Simplified, it's about the ratios of molecules that react with each other. For example, consider the reaction where solid chromium (Cr) reacts with sulfur (S) to form chromium(III) sulfide (Cr₂S₃).

Through stoichiometry, we determine that we need two atoms of Cr for every eight atoms of S to get chromium(III) sulfide. The balanced stoichiometric equation, therefore, shows a 3:1 ratio of sulfur to chromium, emphasizing that chemical reactions depend on the molar ratios of the substances involved.

When students begin to balance a chemical equation, they apply the principles of stoichiometry. As we saw in the solution steps provided, we must first ensure that the number of atoms for each element is the same on both sides of the equation. This not only is vital for proper stoichiometric calculations but also adheres to the law of conservation of mass.

Law of Conservation of Mass

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. It is why when we balance equations, the total mass of reactants must equal the total mass of the products. This principle is fundamental to all chemical equations, including the reactions provided in the exercise.

In the original exercise, to balance the sulfur in the reaction involving chromium and sulfur, we first used a fractional coefficient. However, in finalized chemical equations, we prefer to use whole numbers. Multiplying through by three adjusted our earlier attempt, ensuring that the conservation of mass is respected, and resulted in whole numbers, which are easier to interpret and work with in subsequent calculations.

Maintaining mass in a chemical reaction requires that students balance the equation carefully, considering each element separately. This aspect is critically important because it provides a clear picture that matter transforms rather than vanishes, a concept that was pivotal in the development of chemistry.

Chemical Reaction

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, known as products. They can be simple, involving only a few molecules, or complex, with many substances participating in elaborate sequences of events.

For instance, when sodium bicarbonate is heated, it decomposes into sodium carbonate, carbon dioxide, and water vapor. This type of reaction is a decomposition reaction, which is one of the many types of chemical reactions.

Understanding the nuances of each chemical reaction is vital for mastering balancing techniques. Each reaction has specific ratios in which the reactants combine to form products, and recognizing these ratios helps in balancing the chemical equation. The provided step-by-step solutions display this methodically, breaking down the process to ensure a clear understanding of the necessary stoichiometric balance for each element involved. This comprehensive approach aids students in grasping the underlying principles governing the reactions they’re working with.