Problem 66 - Chemistry The Central Science In Si Units

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

(a) \(\mathrm{CO}_{2} + \mathrm{OH}^{-} \rightarrow \mathrm{HCO}_{3}^{-}\); (b) \(\mathrm{NaHCO}_{3} + \mathrm{H}^{+} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_2\mathrm{O} + \mathrm{Na}^{+}\); (c) \(2\mathrm{CaO} + 5\mathrm{C} \rightarrow 2\mathrm{CaC}_{2} + \mathrm{CO}\); (d) \(\mathrm{C} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{CO} + \mathrm{H}_2\); (e) \(\mathrm{CuO} + \mathrm{CO} \rightarrow \mathrm{Cu} + \mathrm{CO}_{2}\).

Step 1: Complete the First Equation

For equation (a), the reaction between carbon dioxide \(\mathrm{CO}_{2}(g)\) and hydroxide ions \(\mathrm{OH}^{-}(aq)\) forms bicarbonate ions \(\mathrm{HCO}_{3}^{-}(aq)\). Therefore, the complete equation is: \[\mathrm{CO}_{2}(g) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{HCO}_{3}^{-}(aq)\] The equation is already balanced as-is.

Step 2: Complete and Balance the Second Equation

For equation (b), sodium bicarbonate \(\mathrm{NaHCO}_{3}(s)\) reacts with hydrogen ions \(\mathrm{H}^{+}(aq)\) to produce carbon dioxide \(\mathrm{CO}_{2}(g)\), water \(\mathrm{H}_2\mathrm{O}(l)\), and sodium ions \(\mathrm{Na}^{+}(aq)\). The balanced equation is: \[\mathrm{NaHCO}_{3}(s) + \mathrm{H}^{+}(aq) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_2\mathrm{O}(l) + \mathrm{Na}^{+}(aq)\] The equation is balanced as each type of atom appears in the same quantity on both sides.

Step 3: Complete and Balance the Third Equation

For equation (c), calcium oxide \(\mathrm{CaO}(s)\) and carbon \(\mathrm{C}(s)\) react upon heating to form calcium carbide \(\mathrm{CaC}_{2}(s)\) and carbon monoxide \(\mathrm{CO}(g)\). The balanced equation is: \[2\mathrm{CaO}(s) + 5\mathrm{C}(s) \rightarrow 2\mathrm{CaC}_{2}(s) + \mathrm{CO}(g)\] Both sides contain the same number of each type of atom, ensuring balance.

Step 4: Complete and Balance the Fourth Equation

In equation (d), carbon \(\mathrm{C}(s)\) reacts with steam \(\mathrm{H}_{2}\mathrm{O}(g)\) at elevated temperatures to produce carbon monoxide \(\mathrm{CO}(g)\) and hydrogen gas \(\mathrm{H}_2(g)\). The balanced reaction is: \[\mathrm{C}(s) + \mathrm{H}_{2}\mathrm{O}(g) \rightarrow \mathrm{CO}(g) + \mathrm{H}_2(g)\] This equation is balanced as each reactant and product has equivalent numbers of each type of atom.

Step 5: Complete and Balance the Fifth Equation

For equation (e), copper(II) oxide \(\mathrm{CuO}(s)\) reacts with carbon monoxide \(\mathrm{CO}(g)\) to yield copper \(\mathrm{Cu}(s)\) and carbon dioxide \(\mathrm{CO}_{2}(g)\). The finished equation is as follows: \[\mathrm{CuO}(s) + \mathrm{CO}(g) \rightarrow \mathrm{Cu}(s) + \mathrm{CO}_{2}(g)\] This equation maintains balance because each element is conserved on both sides.

Key Concepts

Balancing Equations

Balancing chemical equations is a vital skill in chemistry that ensures we have equal amounts of each type of atom on both sides of the equation. This obeys the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. To balance an equation, you adjust the coefficients, which are the numbers in front of the formulas in the equation, until the number of atoms of each element is the same on both sides. Here's how to go about it:

• First, write down the unbalanced equation.
• Count the number of atoms of each element on both sides.
• Start by balancing elements that appear in only one reactant and one product.
• Next, balance the elements that appear in multiple reactants or products.
• Finally, make sure all coefficients are in the lowest possible ratio.

Balancing can sometimes be challenging, but practicing with different equations can sharpen this crucial skill.

Reaction Stoichiometry

Understanding reaction stoichiometry involves comprehending the quantitative relationships between reactants and products in a chemical reaction. By knowing the balanced equation, we can predict how much of each substance will be needed or produced. Stoichiometry relies heavily on the mole concept. A mole of a substance contains Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\) molecules or atoms. For example, if you have a balanced equation like \(2\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{H}_{2}\mathrm{O}\), it tells us:

• Two moles of hydrogen gas react with one mole of oxygen gas.
• To produce two moles of water vapor.

Through stoichiometry, if you know the amount of one reactant, you can calculate the amount of other reactants needed and products formed, using the ratios given by the balanced equation.

Chemical Reactions

Chemical reactions are processes where substances known as reactants are transformed into different substances called products. They can be classified into several types based on which occurs during the process:

• Synthesis: Two or more reactants combine to form a new product. Example: \(2\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{H}_{2}\mathrm{O}\)
• Decomposition: A single compound breaks down into two or more products. Example: \(2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{H}_{2} + \mathrm{O}_{2}\)
• Single Replacement: An element in a compound is replaced by another element. Example: \(\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2} + \mathrm{H}_{2}\)
• Double Replacement: Exchange of ions between two compounds. Example: \(\mathrm{AgNO}_{3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO}_{3}\)
• Combustion: Oxygen combines with another compound to form water and carbon dioxide. Example: \(\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\)

Each reaction involves a rearrangement of atoms, and understanding these types helps in predicting and mastering chemical processes.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are chemical processes in which electrons are transferred between substances. This transfer changes the oxidation states of the involved elements. An easy way to remember redox processes is through the mnemonic "OIL RIG": Oxidation Is Loss, Reduction Is Gain.- **Oxidation** occurs when a substance loses electrons, leading to an increase in its oxidation state.- **Reduction** takes place when a substance gains electrons, causing a decrease in its oxidation state.Consider the reaction \(\mathrm{CuO} + \mathrm{CO} \rightarrow \mathrm{Cu} + \mathrm{CO}_2\):

• The copper(II) oxide \(\mathrm{CuO}\) is reduced to copper \(\mathrm{Cu}\) by gaining electrons.
• The carbon monoxide \(\mathrm{CO}\) is oxidized to carbon dioxide \(\mathrm{CO}_{2}\) by losing electrons.

Redox reactions are important in many areas, including energy production, corrosion, and biochemical processes. They help understand the electron transfer processes that power both modern technologies and biological systems.