Problem 11 - Chemistry The Central Science

Question

Balance the following equations: $$ \begin{array}{l}{\text { (a) } \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)} \\ {\text { (b) } \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)} \\ {\text { (c) } \mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)} \\ {\text { (d) } \mathrm{Zn}(\mathrm{OH})_{2}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

Step-by-Step Solution

Answer

The balanced equations are: (a) \(\mathrm{CO}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{N}_{2}\mathrm{O}_{5}(g)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{HNO}_{3}(aq)\) (c) \(\mathrm{CH}_{4}(g)+4\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+4\mathrm{HCl}(g)\) (d) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+2\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\)

Step 1: (a) Balancing CO(g) + O2(g) -> CO2(g) equation

1. Check the number of atoms for each element on both sides of the equation: 1 C atom, 2 O atoms on the left, and 1 C atom, 2 O atoms on the right. 2. Since the number of atoms for each element is already equal on both sides, this equation is already balanced: \(\mathrm{CO}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\)

Step 2: (b) Balancing N2O5(g) + H2O(l) -> HNO3(aq) equation

1. Check the number of atoms for each element on both sides of the equation: 2 N atoms, 5 O atoms, and 2 H atoms on the left, and 1 N atom, 3 O atoms, and 1 H atom on the right. 2. Balance N atoms by adding a coefficient of 2 to HNO3(aq): \(\mathrm{N}_{2}\mathrm{O}_{5}(g)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{HNO}_{3}(aq)\) 3. Now we have: 2 N atoms, 5 O atoms, and 2 H atoms on the left, and 2 N atoms, 6 O atoms, and 2 H atoms on the right. 4. Balance the O atoms by adding a coefficient of 2 to H2O(l): \(\mathrm{N}_{2}\mathrm{O}_{5}(g)+2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{HNO}_{3}(aq)\)

Step 3: (c) Balancing CH4(g) + Cl2(g) -> CCl4(l) + HCl(g) equation

1. Check the number of atoms for each element on both sides of the equation: 1 C atom, 4 H atoms, and 2 Cl atoms on the left, and 1 C atom, 4 Cl atoms, and 1 H atom on the right. 2. Balance H atoms by adding a coefficient of 4 to HCl(g): \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+4\mathrm{HCl}(g)\) 3. Now we have: 1 C atom, 4 H atoms, and 2 Cl atoms on the left, and 1 C atom, 4 H atoms, and 8 Cl atoms on the right. 4. Balance the Cl atoms by adding a coefficient of 4 to Cl2(g): \(\mathrm{CH}_{4}(g)+4\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+4\mathrm{HCl}(g)\)

Step 4: (d) Balancing Zn(OH)2(s) + HNO3(aq) -> Zn(NO3)2(aq) + H2O(l) equation

1. Check the number of atoms for each element on both sides of the equation: 1 Zn atom, 2 O atoms, 2 H atoms, 1 N atom on the left, and 1 Zn atom, 6 O atoms, 2 H atoms, 2 N atoms on the right. 2. Balance N atoms by adding a coefficient of 2 to HNO3(aq): \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+2\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(l)\) 3. Now we have: 1 Zn atom, 2 O atoms, 2 H atoms, 2 N atoms on the left, and 1 Zn atom, 6 O atoms, 2 H atoms, and 2 N atoms on the right. 4. Balance the O atoms by adding a coefficient of 2 to H2O(l): \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+2\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)\)

Key Concepts

Stoichiometry

Stoichiometry is like the recipe book for chemistry, allowing chemists to understand and manipulate the quantities involved in chemical reactions. At its core, stoichiometry is about using balanced chemical equations to determine the proportions of reactants and products. This ensures that the mass and charge are conserved throughout the reaction.

• It relies heavily on the concept of the mole, a unit that measures quantity in chemistry, allowing chemists to discuss amounts of substances in terms we can grasp more easily.
• In stoichiometry, coefficients in a balanced equation are used to determine the ratios in which chemicals react or are produced.

For instance, in the example reaction of carbon monoxide combining with oxygen to form carbon dioxide, the coefficients tell us that 1 mole of CO reacts with 1 mole of O₂ to produce 1 mole of CO₂. This is fundamental to ensuring that every element is accounted for, and no atoms mysteriously appear or disappear in the process.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, change into new substances called products. During a chemical reaction, molecules break apart and their atoms rearrange to form different molecules. Understanding chemical reactions is crucial as they are at the heart of everything from digesting food to powering engines.

• Chemical reactions can be exothermic, releasing energy, or endothermic, absorbing energy.
• Balancing chemical reactions ensures the conservation of matter, adhering to the law of conservation of mass.

In our examples, the reaction of Zn(OH)₂ with HNO₃ illustrates a chemical reaction involving complex ion exchange, resulting in new compounds and mixtures.

Molecular Formulas

Molecular formulas provide essential information about a chemical compound. They show the exact number and type of atoms present in a molecule. These formulas are akin to a snapshot of a molecule’s composition, giving us insight into the structure, although not necessarily the arrangement of atoms.

• For instance, the molecular formula of water, H₂O, tells us that each molecule contains two hydrogen atoms and one oxygen atom.
• Molecular formulas are crucial for understanding and predicting chemical behavior and properties.

The ability to decipher these formulas allows chemists to recognize the fundamental architecture of compounds, predict outcomes in reactions, and synthesize new materials with desired traits. For chemical reactions to be balanced like in our exercises, the correct interpretation of these molecular formulas is vital, as every atom in the reactants must find its place in the resulting products.