Problem 13 - Chemistry The Central Science In Si Units

Question

Balance the following equations: (a) \(\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)\) (d) \(\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)\)

Step-by-Step Solution

Answer

Balanced equations are as follows: (a) \(2\mathrm{CaS}+2\mathrm{H}_{2}\mathrm{O}\rightarrow \mathrm{Ca}(\mathrm{HS})_{2}+\mathrm{Ca}(\mathrm{OH})_{2}\) (b) \(4\mathrm{NH}_{3}+5\mathrm{O}_{2}\rightarrow 4\mathrm{NO}+6\mathrm{H}_{2}\mathrm{O}\) (c) \(2\mathrm{FeCl}_{3}+3\mathrm{Na}_{2}\mathrm{CO}_{3}\rightarrow \mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3}+6\mathrm{NaCl}\) (d) \(2\mathrm{FeS}_{2}+7\mathrm{O}_{2}\rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}+4\mathrm{SO}_{2}\).

Step 1: Analyze the Equation (a)

Consider the given skeleton equation: \(\mathrm{CaS}(s)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(aq)+\mathrm{Ca}(\mathrm{OH})_{2}(aq)\). Identify elements to balance: Ca, S, O, H.

Step 2: Balance Calcium (Ca) in Equation (a)

The skeleton equation has 1 Ca atom on each side. Thus, calcium is already balanced.

Step 3: Balance Sulfur (S) in Equation (a)

There is 1 S atom on the left and 2 S atoms in \(\mathrm{Ca}(\mathrm{HS})_{2}\). Multiply \(\mathrm{CaS}\) by 2: \(2\mathrm{CaS}(s)\).

Step 4: Update and Balance Hydrogen (H) and Oxygen (O) in Equation (a)

Now add 2 water molecules to balance hydrogen: \(2\mathrm{H}_{2}\mathrm{O}\). For oxygen, both sides now have 2 oxygen atoms, achieving a balance.

Step 5: Verify Balanced Equation (a)

Equation becomes \(2\mathrm{CaS}(s)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(aq)+\mathrm{Ca}(\mathrm{OH})_{2}(aq)\).

Step 6: Analyze the Equation (b)

For \(\mathrm{NH}_{3}+\mathrm{O}_{2}\longrightarrow\mathrm{NO}+\mathrm{H}_{2}\mathrm{O}\), identify elements for balancing: N, O, H.

Step 7: Balance Nitrogen (N) in Equation (b)

There is 1 N on the left and 1 N atom on the right. Nitrogen is balanced.

Step 8: Balance Hydrogen (H) in Equation (b)

There are 3 H atoms on the left in NH\(_3\). On the right, it contributes to forming 3/2 \(\mathrm{H}_2\mathrm{O}\). Multiply water by a fraction 1.5: \(1.5\mathrm{H}_2\mathrm{O}\) then double the entire equation to remove fractions.

Step 9: Balance Oxygen (O) in Equation (b)

There are 3 O atoms in \(1.5\mathrm{H}_2\mathrm{O}\) and \(\mathrm{O}_{2}\). Double the reaction: \(4\mathrm{NH}_{3}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_{2}\mathrm{O}(g)\).

Step 10: Verify Balanced Equation (b)

The balanced equation is: \(4\mathrm{NH}_{3}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_{2}\mathrm{O}(g)\).

Step 11: Analyze the Equation (c)

For \(\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(aq) \longrightarrow \mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3}(s)+\mathrm{NaCl}(aq)\), identify atoms: Fe, Cl, Na, C, O.

Step 12: Balance Iron (Fe) in Equation (c)

There are 2 Fe atoms in \(\mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3}\). Multiply \(\mathrm{FeCl}_{3}\) by 2: \(2\mathrm{FeCl}_{3}\).

Step 13: Balance Sodium (Na) and Chlorine (Cl) in Equation (c)

There are 6 Na on right. Multiply \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) by 3 to balance sodium: \(3\mathrm{Na}_{2}\mathrm{CO}_{3}\). Chlorine balances naturally at 6.

Step 14: Verify Balanced Equation (c)

Balanced equation: \(2\mathrm{FeCl}_{3}(s) + 3\mathrm{Na}_{2}\mathrm{CO}_{3}(aq) \longrightarrow \mathrm{Fe}_{2}(\mathrm{CO}_{3})_{3}(s) + 6\mathrm{NaCl}(aq)\).

Step 15: Analyze the Equation (d)

Consider \(\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g)\longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)\). Identify: Fe, S, O.

Step 16: Balance Iron (Fe) in Equation (d)

2 Fe in \(\mathrm{Fe}_{2}\mathrm{O}_{3}\). Multiply \(\mathrm{FeS}_{2}\) by 2: \(2\mathrm{FeS}_{2}(s)\).

Step 17: Balance Sulfur (S) in Equation (d)

There are 4 S on the left (from 2\(\mathrm{FeS}_{2}\)), thus need 4 \(\mathrm{SO}_{2}\).

Step 18: Balance Oxygen (O) in Equation (d)

Balancing oxygen: right side has 6 O in \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) and 8 in \(\mathrm{SO}_{2}\). Total: 14 O atoms, requiring 7 \(\mathrm{O}_{2}\): \(2\mathrm{FeS}_{2}(s)+7\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)+4\mathrm{SO}_{2}(g)\).

Step 19: Verify Balanced Equation (d)

Balanced equation: \(2\mathrm{FeS}_{2}(s)+7\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)+4\mathrm{SO}_{2}(g)\).

Key Concepts

Stoichiometry

Stoichiometry is the backbone of balancing chemical equations. It's like a recipe, ensuring all ingredients (or reactants) perfectly form the desired products without anything leftover. When you're dealing with stoichiometry, think of it as a system of accounting for atoms. Every atom present on the reactant side must also appear on the product side.

The main aspects involve:

• Identifying the reaction components and their respective formulas.
• Determining the number of atoms for each element involved in the reaction.
• Altering coefficients to achieve balance, ensuring that both sides of the equation have an equal number of each type of atom.

Balancing equations with stoichiometry requires you to adjust the molecules but never the subscripts in a compound. This adjustment affects how many of each molecule participate in a reaction. For example, in equation (a) from the exercise, we begin with one calcium sulfide and need to adjust it to two to balance sulfur across the equation. It's a methodical approach to make sure nothing is created or destroyed in a chemical reaction, aligning perfectly with the law of conservation of mass.

Chemical Reactions

Chemical reactions are processes where reactants transform into products. It's a change that’s observed through different signs such as color change, temperature change, gas production, or precipitate formation. In the context of these exercises, we're dealing with several types of reactions:

• Combination reactions, where two or more reactants form one product.
• Decomposition reactions, where a compound breaks down into two or more simpler substances.
• Single replacement reactions, where an element replaces another in a compound.
• Double replacement reactions, where there is an exchange of ions between two compounds resulting in different products.

In equation (c), we see a double replacement reaction happening where sodium carbonate reacts with iron chloride. Understanding the type of reaction occurring helps in predicting the products and balancing the equation correctly. In essence, knowing these reactions allows chemists to anticipate what happens when substances interact and to control the outcomes for practical applications.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often called redox reactions, involve the transfer of electrons between two substances. These reactions are common in many chemical processes, including those in biological, terrestrial, and industrial contexts. Understanding these involves recognizing the oxidation state changes in substances as they react.

Key points to consider:

• Oxidation refers to the loss of electrons, resulting in an increase in oxidation state.
• Reduction refers to the gain of electrons, resulting in a decrease in oxidation state.
• The substance that donates electrons becomes oxidized, and the one that accepts electrons becomes reduced.

An example from the exercises is in equation (d), where iron sulfide reacts with oxygen, leading to iron oxide and sulfur dioxide. In this redox process, iron changes oxidation states, exemplifying the transfer of electrons fueling these reactions. Understanding redox reactions is crucial in fields ranging from energy storage to metabolic pathways in biology. It highlights the importance of electron exchange in sustaining chemical change.