Question

Complete and balance each equation: (a) $\mathrm{PbS}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{⟶}$ (b) ${\mathrm{CoCO}}_3(\mathrm{s})\stackrel{\mathrm{\Delta }}{⟶}$ (c) ${\mathrm{WO}}_3(\mathrm{s})+{\mathrm{H}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{⟶}$ (d) ${\mathrm{VCl}}_3(g)+\mathrm{K}(I)⟶$ (e) $\mathrm{BaO}(s)+{\mathrm{P}}_2{\mathrm{O}}_5(l)⟶$

Short answer

(a) $PbS(s)+O_2(g)\to PbS{O}_4(s)$ (b) $CoC{O}_3(s)\to CoO(s)+C{O}_2(g)$ (c) $W{O}_3(s)+2H_2(g)\to W{O}_2(s)+H_{2}O(l)$ (d) $2VC{l}_3(g)+3KI\to 2VC{l}_2(s)+3KCl(s)+I_2(s)$ (e) $3BaO(s)+P_2{O}_5(l)\to B{a}_3(P{O}_4{)}_2(s)$

Step-by-step solution

(a) Balancing PbS(s) + O₂(g) ⟶/tag_title#

To balance this equation, compare the number of atoms in the reactant side and the product side. We have 1 Pb atom, 1 S atom and 2 O atoms on the reactant side. Since we have O₂ on the reactant side, the product must also contain oxygen. In this case, it will be the corresponding oxide of Pb and S: PbSO₄ Now, write the unbalanced equation: PbS(s) + O₂(g) ⟶ PbSO₄(s) To balance the equation, notice that the number of S and O atoms are already balanced, so no further adjustments are needed. The balanced equation is: PbS(s) + O₂(g) ⟶ PbSO₄(s)

(b) Balancing CoCO₃(s) ⟶/tag_title#

Observe the given reactant contains cobalt (Co) and carbonate (CO₃). Upon heating, the carbonate will break down releasing carbon dioxide (CO₂) and leaving the corresponding oxide, which is cobalt(II) oxide (CoO). Write the unbalanced equation: CoCO₃(s) ⟶ CoO(s) + CO₂(g) In this case, the equation is already balanced with one cobalt, one carbon, and three oxygen atoms on each side.

(c) Balancing WO₃(s) + H₂(g) ⟶/tag_title#

On the reactant side, we have tungsten trioxide (WO₃) and hydrogen gas (H₂), meaning we have 1 tungsten, 3 oxygen, and 2 hydrogen atoms. Since there's hydrogen involved and an oxide of tungsten present, the most probable product will be the reduction of tungsten trioxide to a lower oxide and water: WO₃(s) + H₂(g) ⟶ WO₂(s) + H₂O(l) To balance this equation, we need 2 hydrogen atoms in the product side for water. Thus, multiply H₂(g) by 2: WO₃(s) + 2H₂(g) ⟶ WO₂(s) + H₂O(l) Now the equation is balanced.

(d) Balancing VCl₃(g) + KI ⟶/tag_title#

Here, we have vanadium trichloride (VCl₃) reacting with potassium iodide (KI). This is most probably a redox/oxidation-reduction reaction, where there will be an exchange of anions: VCl₃(g) + KI ⟶ VCl₂(s) + KCl(s) + I₂(s) To balance this equation, first balance the iodide (I): 2VCl₃(g) + 3KI ⟶ 2VCl₂(s) + 3KCl(s) + I₂(s) Now we have 2 vanadium, 6 chlorine, 3 potassium, and 2 iodine atoms on each side, so the equation is balanced.

(e) Balancing BaO(s) + P₂O₅(l) ⟶/tag_title#

In this equation, we have barium oxide (BaO) reacting with phosphorus pentoxide (P₂O₅). This reaction will form an ionic compound, where barium (Ba²⁺) will form a bond with phosphate (PO₄³⁻) ions forming barium phosphate: BaO(s) + P₂O₅(l) ⟶ Ba₃(PO₄)₂(s) To balance the equation, we need to have 3 barium, 2 phosphorus, and 8 oxygen atoms on each side. Therefore, adjust the coefficients like this: 3BaO(s) + P₂O₅(l) ⟶ Ba₃(PO₄)₂(s) Now, the equation is balanced.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the elements and compounds as they undergo chemical transformations. To comprehend stoichiometry, one must understand the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This law is the foundation for balancing chemical equations, ensuring that the number of atoms for each element is the same on both the reactants' and products' sides.

For instance, when balancing the reaction between lead sulfide (PbS) and oxygen (O₂), which produces lead sulfate (PbSO₄), stoichiometry is used to ensure the atoms are balanced, signifying we have equivalent amounts of lead (Pb), sulfur (S), and oxygen (O) before and after the reaction. Analyzing stoichiometric coefficients in a balanced chemical equation allows us to predict the quantities of reactants required and products formed, making it a critical concept in chemistry.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are chemical reactions that involve the transfer of electrons between two species. An oxidation reaction involves the loss of electrons, whereas reduction involves the gain of electrons. Redox reactions are common in nature and technology, like in batteries and metabolic processes.

In the reaction where vanadium trichloride (VCl₃) reacts with potassium iodide (KI), it's the movement of electrons that changes the oxidation states of the elements, leading to a balanced redox equation. It's crucial to identify the elements getting oxidized and reduced to correctly balance redox reactions. This is often achieved by assigning oxidation numbers to each element in the reaction and ensuring that the total change in oxidation numbers for each element is compensated by an opposite change in another.

Chemical Reaction Types

Chemical reactions are categorized into types based on their characteristics and the changes that the reactants undergo to form products. Common types of chemical reactions include synthesis, decomposition, single displacement, double displacement, combustion, and redox reactions. Each type has specific patterns that help predict the products of a reaction.

As an exercise in identifying reaction types, consider the decomposition of cobalt(II) carbonate (CoCO₃) upon heating, which produces cobalt(II) oxide (CoO) and carbon dioxide (CO₂). This reaction is classified as a decomposition reaction because one compound breaks down into two or more simpler substances. Understanding these types of reactions assists in inferring possible products and aids in the process of balancing the chemical equations. It is this understanding of the different reaction types that can guide students towards an improved comprehension and application of chemical equation balancing.