Problem 68 - Chemistry The Central Science

Question

Complete and balance the following equations: (a) \(\mathrm{CO}_{2}(g)+\mathrm{OH}^{-}(a q) \longrightarrow\) (b) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{CaO}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\Delta}{\longrightarrow}\) (e) \(\mathrm{CuO}(s)+\mathrm{CO}(g) \longrightarrow\)

Step-by-Step Solution

Answer

The short answer is: (a) CO₂(g) + OH⁻(aq) → HCO₃⁻(aq) (b) NaHCO₃(s) + H⁺(aq) → Na⁺(aq) + CO₂(g) + H₂O(l) (c) CaO(s) + 2 C(s) → CaC₂(s) (d) C(s) + H₂O(g) → CO(g) + H₂(g) (e) CuO(s) + CO(g) → Cu(s) + CO₂(g)

Step 1: (Identify Products)

For the first reaction, CO₂(g) + OH⁻(aq) react together. The possible product of this reaction is the formation of bicarbonate ion, HCO₃⁻(aq) since hydrogen from OH⁻ would bond to CO₂. \ #a) Balancing the Reaction#

Step 2: (Balance Atoms)

To balance the equation, adjust the coefficients: \( \) 1 CO₂(g) + 1 OH⁻(aq) → 1 HCO₃⁻(aq) \ Here, the number of atoms for each element is the same on both sides of the equation. Thus, the balanced equation is: \( \) CO₂(g) + OH⁻(aq) → HCO₃⁻(aq) #b) Completing the Reaction#

Step 3: (Identify Products)

For the second reaction, NaHCO₃(s) + H⁺(aq) react together. The possible products of this reaction are Na⁺(aq), CO₂(g), and H₂O(l) because the H⁺ would bond with the HCO₃⁻ to form water and release carbon dioxide. \ #b) Balancing the Reaction#

Step 4: (Balance Atoms)

To balance the equation, adjust the coefficients: \( \) 1 NaHCO₃(s) + 1 H⁺(aq) → 1 Na⁺(aq) + 1 CO₂(g) + 1 H₂O(l) \ Here, the number of atoms for each element is the same on both sides of the equation. Thus, the balanced equation is: \( \) NaHCO₃(s) + H⁺(aq) → Na⁺(aq) + CO₂(g) + H₂O(l) #c) Completing the Reaction#

Step 5: (Identify Products)

For the third reaction, CaO(s) + C(s) react together with Δ (heat). The possible product of this reaction is the formation of calcium carbide, CaC₂(s), because calcium would bond with carbon. \( \) #c) Balancing the Reaction#

Step 6: (Balance Atoms)

To balance the equation, adjust the coefficients: \( \) 1 CaO(s) + 2 C(s) → 1 CaC₂(s) \ Here, the number of atoms for each element is the same on both sides of the equation. Thus, the balanced equation is: \( \) CaO(s) + 2 C(s) → CaC₂(s) #d) Completing the Reaction#

Step 7: (Identify Products)

For the fourth reaction, C(s) + H₂O(g) react together with Δ (heat). The possible product of this reaction is the formation of carbon monoxide, CO(g) and hydrogen gas, H₂(g) since carbon would bond with oxygen and hydrogen is released as a gas. \( \) #d) Balancing the Reaction#

Step 8: (Balance Atoms)

To balance the equation, adjust the coefficients: \( \) 1 C(s) + 1 H₂O(g) → 1 CO(g) + 1 H₂(g) \ Here, the number of atoms for each element is the same on both sides of the equation. Thus, the balanced equation is: \( \) C(s) + H₂O(g) → CO(g) + H₂(g) #e) Completing the Reaction#

Step 9: (Identify Products)

For the fifth reaction, CuO(s) + CO(g) react together. The possible product of this reaction is the formation of copper, Cu(s) and carbon dioxide, CO₂(g) as carbon would bond with the oxygen bound to copper. \ #e) Balancing the Reaction#

Step 10: (Balance Atoms)

To balance the equation, adjust the coefficients: \( \) 1 CuO(s) + 1 CO(g) → 1 Cu(s) + 1 CO₂(g) \ Here, the number of atoms for each element is the same on both sides of the equation. Thus, the balanced equation is: \( \) CuO(s) + CO(g) → Cu(s) + CO₂(g)

Key Concepts

Equation Balancing

Balancing a chemical equation ensures that the number of atoms for each element is the same on both sides of the equation. This reflects the conservation of mass, where matter cannot be created or destroyed in a chemical reaction. To balance an equation, start by counting the number of atoms of each element in the reactants and products. Then, adjust the coefficients, which are the numbers placed before compounds, to make the number of atoms equal. - Begin with the most complex molecule, often the one with the greatest number of different atoms.- Adjust coefficients to achieve the same atom count for each element on both sides.- Always check your final equation to ensure that both mass and charge are balanced, especially in reactions involving ions.For example, in the reaction between carbon dioxide (\(\text{CO}_2\)) and hydroxide (\(\text{OH}^-\)) the reaction forms the bicarbonate ion (\(\text{HCO}_3^-\)). Coordinating the coefficients maintains the balance of atoms for carbon, oxygen, and hydrogen. The balanced equation is:\[\text{CO}_2(g) + \text{OH}^-(aq) \rightarrow \text{HCO}_3^-(aq)\]

Reactants and Products

In a chemical reaction, substances called reactants transform into substances called products. Understanding what happens at the atomic level can make recognizing these changes easier.- **Reactants** are the starting materials in a reaction, generally shown on the left side of the equation.- **Products** are the compounds formed as a result of the reaction, shown on the right end of the equation.Each chemical equation tells a unique story. For instance, consider the reaction of CuO(s) with CO(g). Here, copper(II) oxide and carbon monoxide are the reactants that react to form copper and carbon dioxide, the new products:\[\text{CuO}(s) + \text{CO}(g) \rightarrow \text{Cu}(s) + \text{CO}_2(g)\]Notice how atoms rearrange during the reaction, yet still preserve the type and number of atoms between the reactants and products.

Stoichiometry

Stoichiometry plays a key role in understanding the relationships between reactants and products in a chemical reaction. It involves the calculation of the relative quantities of reactants and products.Key aspects of stoichiometry include:- **Moles**: A mole is a unit that measures the amount of substance. Stoichiometry allows you to convert between grams and moles using the molecular weights from the periodic table. - **Ratios**: The coefficients in a balanced equation indicate stoichiometric ratios, showing how much of one substance reacts with another.In the case of the reaction between NaHCO₃ and H⁺, stoichiometry tells us that one mole of sodium bicarbonate reacts with one mole of hydrogen ions to produce one mole each of sodium, carbon dioxide, and water:\[\text{NaHCO}_3(s) + \text{H}^+(aq) \rightarrow \text{Na}^+(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\]These calculated and experimental stoichiometric relationships are crucial for scaling reactions, whether for laboratory experiments or industrial chemical processes.