Question

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

Short answer

The short answers are: a) \( CO_{2}(g) + OH^-(aq) \longrightarrow H_{2}O(l) + HCO_{3}^-(aq) \) b) \( 2 NaHCO_{3}(s) + H^{+}(aq) \longrightarrow H_{2}O(l) + Na_{2}CO_{3}(s) \) c) \( CaO(s) + C(s) \stackrel{\Delta}{\longrightarrow} CaC_{2}(s) + CO(g) \) d) \( C(s) + H_{2}O(g) \stackrel{\Delta}{\longrightarrow} CO(g) + H_{2}(g) \) e) \( CuO(s) + CO(g) \longrightarrow Cu(s) + CO_{2}(g) \)

Step-by-step solution

Identify the products

In this reaction, carbon dioxide (CO₂) reacts with hydroxide (OH⁻) ions. They will likely form water (H₂O) and a bicarbonate anion (HCO₃⁻). $$ \mathrm{CO}_{2}(g) + \mathrm{OH}^-(aq) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{HCO}_{3}^-(aq) $$

Balance the equation

The equation seems balanced since there are equal numbers of carbon, oxygen, and hydrogen atoms on both sides of the equation. $$ \mathrm{CO}_{2}(g) + \mathrm{OH}^-(aq) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{HCO}_{3}^-(aq) $$ #b) NaHCO₃(s) + H⁺(aq) → #

Identify the products

In this reaction, sodium bicarbonate (NaHCO₃) reacts with a hydrogen ion (H⁺). They will form water (H₂O) and sodium carbonate (Na₂CO₃). $$ \mathrm{NaHCO}_{3}(s) + \mathrm{H}^{+}(aq) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Na}_{2}\mathrm{CO}_{3}(s) $$

Balance the equation

Since there are two sodium atoms on the product side, we need to double the amount of sodium bicarbonate on the reactant side: $$ 2 \mathrm{NaHCO}_{3}(s) + \mathrm{H}^{+}(aq) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Na}_{2}\mathrm{CO}_{3}(s) $$ #c) CaO(s) + C(s) → #

Identify the products

In this reaction, calcium oxide (CaO) reacts with carbon (C) in the presence of heat. The products form calcium carbide (CaC₃) and carbon monoxide (CO). $$ \mathrm{CaO}(s) + \mathrm{C}(s) \stackrel{\Delta}{\longrightarrow} \mathrm{CaC}_{2}(s) + \mathrm{CO}(g) $$

Balance the Equation

This equation is already balanced. $$ \mathrm{CaO}(s) + \mathrm{C}(s) \stackrel{\Delta}{\longrightarrow} \mathrm{CaC}_{2}(s) + \mathrm{CO}(g) $$ #d) C(s) + H₂O(g) → #

Identify the products

In this reaction, carbon (C) reacts with water (H₂O) in the presence of heat. The products formed will be carbon monoxide (CO) and hydrogen gas (H₂). $$ \mathrm{C}(s) + \mathrm{H}_{2}\mathrm{O}(g) \stackrel{\Delta}{\longrightarrow} \mathrm{CO}(g) + \mathrm{H}_{2}(g) $$

Balance the Equation

This equation is already balanced. $$ \mathrm{C}(s) + \mathrm{H}_{2}\mathrm{O}(g) \stackrel{\Delta}{\longrightarrow} \mathrm{CO}(g) + \mathrm{H}_{2}(g) $$ #e) CuO(s) + CO(g) → #

Identify the products

In this reaction, copper oxide (CuO) reacts with carbon monoxide (CO), forming copper (Cu) and carbon dioxide (CO₂). $$ \mathrm{CuO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Cu}(s) + \mathrm{CO}_{2}(g) $$

Balance the Equation

This equation is already balanced. $$ \mathrm{CuO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Cu}(s) + \mathrm{CO}_{2}(g) $$

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in chemical reactions. It is the calculation of the quantities of chemical elements or compounds involved in chemical reactions. Understanding stoichiometry is crucial because it ensures that the law of conservation of mass is upheld, meaning matter is not created or destroyed in a chemical reaction.

For example, consider the balanced chemical equation from the exercise:$$2 \text{ NaHCO}_{3}(s) + \text{H}^{+}(aq) \rightarrow \text{H}_{2}\text{O}(l) + \text{Na}_{2}\text{CO}_{3}(s)$$This means that two moles of sodium bicarbonate react with one mole of hydrogen ions to produce one mole of water and one mole of sodium carbonate. The coefficients of the compounds in the balanced equation are in a stoichiometric ratio that reflects this precise quantitative relationship.

To balance a chemical equation stoichiometrically, you must ensure that the number of atoms for each element is the same on both sides of the equation. This reflects the conservation of mass, as the total mass of the reactants must equal the total mass of the products.

Chemical Reactions

A chemical reaction involves the transformation of one or more substances into different substances. Changes at the molecular level result in new chemical structures with different properties. Every chemical reaction can be represented by an equation that lists the reactants and products. This graphical representation is crucial for understanding what occurs at the molecular level during a reaction.

In one of the given exercises, copper oxide reacts with carbon monoxide, which is a reduction-oxidation reaction:$$\text{CuO}(s) + \text{CO}(g) \rightarrow \text{Cu}(s) + \text{CO}_{2}(g)$$This equation tells us that each molecule of copper oxide reacts with a molecule of carbon monoxide to yield metallic copper and carbon dioxide gas. The physical states of the components (solid, gas, aqueous) are also indicated, which provides further insight into the reaction conditions.

Law of Conservation of Mass

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This law is the fundamental principle that underlies the balancing of chemical equations. When solving any chemical equation, it's essential to balance it so that the mass of the reactants equals the mass of the products.

For instance, when calcium oxide (CaO) reacts with carbon to form calcium carbide (CaC₂) and carbon monoxide (CO), the original mass of the reactants must equal the final mass of the products:$$\text{CaO}(s) + \text{C}(s) \rightarrow \text{CaC}_{2}(s) + \text{CO}(g)$$In this balanced equation, you can verify the law of conservation of mass by tallying the number of each type of atom on both sides and finding them to be equal. This concept is fundamental to all chemical equations and reactions.