Problem 65 - Chemistry: The Central Science in SI Units, Global Edition

Question

Complete and balance the following equations: (a) \(\mathrm{ZnCO}_{3}(s) \stackrel{\Delta}{\longrightarrow}\) (b) \(\mathrm{BaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) (d) \(\mathrm{CS}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow\) (e) \(\mathrm{Ca}(\mathrm{CN})_{2}(s)+\operatorname{HBr}(a q) \longrightarrow\)

Step-by-Step Solution

Answer

(a) \( ZnCO_3(s) \stackrel{\Delta}{\longrightarrow} ZnO(s) + CO_2(g) \) (b) \( BaC_2(s) + 2 H_2O(l) \longrightarrow Ba(OH)_2(s) + C_2H_2(g) \) (c) \( 2 C_2H_2(g) + 5 O_2(g) \longrightarrow 4 CO_2(g) + 4 H_2O(g) \) (d) \( CS_2(g) + 3 O_2(g) \longrightarrow CO_2(g) + 2 SO_2(g) \) (e) \( Ca(CN)_2(s) + 2 HBr(aq) \longrightarrow CaBr_2(aq) + 2 HCN(aq) \)

Step 1: Balance equation (a)

(a) ZnCO3(s) → ZnO(s) + CO2(g) Since there is 1 Zn, 1 C, and 3 O atoms on both sides of the equation, it is already balanced.

Step 2: Balance equation (b)

(b) BaC2(s) + H2O(l) → Ba(OH)2(s) + C2H2(g) First, balance the Ba atoms: 1 Ba on both sides, no change needed. Next, balance the C atoms: 2 C on the left, 2 C on the right, no change needed. Finally, balance the H and O atoms by placing a "2" in front of H2O and Ba(OH)2: BaC2(s) + 2 H2O(l) → Ba(OH)2(s) + C2H2(g)

Step 3: Balance equation (c)

(c) C2H2(g) + O2(g) → CO2(g) + H2O(g) First, balance the C atoms by placing a "2" in front of CO2: C2H2(g) + O2(g) → 2 CO2(g) + H2O(g) Next, balance the H atoms by placing a "2" in front of H2O: C2H2(g) + O2(g) → 2 CO2(g) + 2 H2O(g) Finally, balance the O atoms by placing a "5/2" in front of O2 and multiplying all coefficients by 2: 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(g)

Step 4: Balance equation (d)

(d) CS2(g) + O2(g) → CO2(g) + SO2(g) First, balance the C atoms: 1 C on both sides, no change needed. Next, balance the S atoms by placing a "2" in front of SO2: CS2(g) + O2(g) → CO2(g) + 2 SO2(g) Finally, balance the O atoms by placing a "3" in front of O2: CS2(g) + 3 O2(g) → CO2(g) + 2 SO2(g)

Step 5: Balance equation (e)

(e) Ca(CN)2(s) + HBr(aq) → CaBr2(aq) + HCN(aq) First, balance the Ca atoms: 1 Ca on both sides, no change needed. Next, balance the Br atoms by placing a "2" in front of HBr and CaBr2: Ca(CN)2(s) + 2 HBr(aq) → CaBr2(aq) + HCN(aq) Finally, balance the C and N atoms by placing a "2" in front of HCN: Ca(CN)2(s) + 2 HBr(aq) → CaBr2(aq) + 2 HCN(aq)

Key Concepts

Balancing Chemical Reactions

To understand balancing chemical reactions, it is crucial to grasp the concept of the law of conservation of mass. This law states that matter is neither created nor destroyed in a chemical reaction. Balancing chemical equations ensures that this law is followed. Each type of atom present in the reactants must be equal to those in the products.

• Start by identifying all reactants and products involved.
• Count the number of each type of atom on both sides of the equation.
• Add coefficients to the compounds in the equation to equalize the number of each type of atom on both sides.
• Adjust and repeat the process as needed until balanced.

By following these steps, you will ensure that your chemical equations reflect the real-world changes occurring during a reaction.

Chemical Decomposition

Chemical decomposition is a type of reaction where a single compound breaks down into two or more simpler substances. This process is usually initiated by heating, voltage, or a catalyst. In the provided exercise, equation (a) represents a decomposition reaction:
ZnCOe needs to heat back up 3(s) \( \longrightarrow \) ZnO(s) + COdisemination2(g).
Key points to remember about decomposition reactions include:

• Typically, they require energy in the form of heat, light, or electricity.
• Often involve a single reactant.
• Produces multiple simpler products like elements or simpler compounds.

Decomposition reactions are prevalent in various fields, such as the decomposition of carbonates to form oxides and carbon dioxide.

Combustion Reactions

Combustion reactions are characterized by the process of burning, where a substance reacts rapidly with oxygen to release energy. These reactions produce heat and light, making them exothermic. In the steps provided, equation (c) illustrates a combustion reaction:
Cdisemination2Hdisemination2(g) + Odisemination2(g) \( \longrightarrow \) COdisemination2(g) + Hdisemination2O(g).
Important features of combustion reactions include:

• Reactants typically include a hydrocarbon and oxygen.
• Products are usually carbon dioxide and water.
• They are a significant source of energy in everyday contexts, such as power plants and engines.

Understanding combustion reactions can help in exploring energy production and initiating further studies on sustainable energy sources.

Stoichiometry

Stoichiometry is the area of chemistry that involves quantitative relationships between reactants and products in a chemical reaction. It is essential for predicting how much of a product will form and reactant will be needed.
To approach stoichiometry, follow these steps:

• Ensure the chemical equation is balanced, as seen in all equations from the exercise.
• Use the balanced equation to derive mole ratios between reactants and products.
• Apply these mole ratios to convert between masses, moles, or even particles of the reactants and products.

Overall, stoichiometry enables chemists to perform practical and calculated predictions, critical in both laboratory and industrial applications, ensuring efficient use of resources.