Question

Complete and balance each equation: (a) \(\mathrm{PbS}(s)+\mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow}\) (b) \(\mathrm{CoCO}_{3}(s) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{WO}_{3}(s)+\mathrm{H}_{2}(g) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{VCl}_{3}(g)+\mathrm{K}(l) \longrightarrow\) (e) \(\mathrm{BaO}(s)+\mathrm{P}_{2} \mathrm{O}_{5}(l) \longrightarrow\)

Short answer

(a) \(PbS(s) + O_2(g) \longrightarrow PbO(s) + SO_2(g)\) (b) \(CoCO_3(s) \stackrel{\Delta}{\longrightarrow} CoO(s) + CO_2(g)\) (c) \(WO_3(s) + 3H_2(g) \stackrel{\Delta}{\longrightarrow} W(s) + 3H_2O(l)\) (d) \(VCl_3(g) + 2K(l) \longrightarrow VKCl_2(s) + KCl(s)\) (e) \(2BaO(s) + P_2O_5(l) \longrightarrow Ba(PO_3)_2(s)\)

Step-by-step solution

(a) Write the unbalanced equation

First, identify the products of the reaction, PbS and O2 react with each other, so the balanced equation should be: \[PbS(s) + O_2(g) \longrightarrow PbO(s) + SO_2(g)\]

(a) Balance the equation

To balance the equation, we need to ensure that there's the same number of atoms of each element on both sides. The equation can be balanced by modifying it to: \[PbS(s) + O_2(g) \longrightarrow PbO(s) + SO_2(g)\]

(b) Write the unbalanced equation

First, identify the products of the reaction, as CoCO3 decomposes: \[CoCO_3(s) \stackrel{\Delta}{\longrightarrow} CoO(s) + CO_2(g)\]

(b) Balance the equation

The equation is already balanced as is: \[CoCO_3(s) \stackrel{\Delta}{\longrightarrow} CoO(s) + CO_2(g)\]

(c) Write the unbalanced equation

First, identify the products of the reaction, WO3 and H2 react with each other: \[WO_3(s) + H_2(g) \stackrel{\Delta}{\longrightarrow} W(s) + H_2O(l)\]

(c) Balance the equation

To balance the equation, add coefficients: \[WO_3(s) + 3H_2(g) \stackrel{\Delta}{\longrightarrow} W(s) + 3H_2O(l)\]

(d) Write the unbalanced equation

First, identify the products of the reaction, VCl3 and K react with each other: \[VCl_3(g) + K(l) \longrightarrow VKCl_2(s) + KCl(s)\]

(d) Balance the equation

To balance the equation, add coefficients: \[VCl_3(g) + 2K(l) \longrightarrow VKCl_2(s) + KCl(s)\]

(e) Write the unbalanced equation

First, identify the products of the reaction, BaO and P2O5 react with each other: \[BaO(s) + P_2O_5(l) \longrightarrow Ba(PO_3)_2(s)\]

(e) Balance the equation

To balance the equation, add coefficients: \[2BaO(s) + P_2O_5(l) \longrightarrow Ba(PO_3)_2(s)\]

Key concepts

Stoichiometry

Stoichiometry is at the heart of chemical reactions, acting as a mathematical approach to understanding and quantifying the relationship between reactants and products in a chemical equation. It's akin to a recipe in cooking: stoichiometry tells you how much of each ingredient you need to end up with a specific amount of the final product.

For students, mastering stoichiometry involves becoming comfortable with the mole concept (the unit that measures the amount of substance), balancing chemical equations, and being able to calculate the masses of reactants and products. When looking at our textbook exercise, pay attention to the coefficients in the balanced equations; they represent the 'mole ratio' and provide the key to stoichiometric calculations. For instance, in step 6:
\[WO_3(s) + 3H_2(g) \rightarrow W(s) + 3H_2O(l)\]
each mole of tungsten(VI) oxide reacts with three moles of hydrogen gas to produce one mole of tungsten and three moles of water. Understanding this numerical relationship is crucial for predicting the yields of the reactions and for performing any consequent stoichiometric calculations.

Chemical Reactions

Chemical reactions are processes where reactants transform into products through the breaking and forming of chemical bonds. They can be categorized based on their nature and the outcome of the reaction. For instance, synthesis or combination reactions involve two or more substances combining to form a new compound, whereas decomposition reactions involve breaking down a compound into simpler substances, often under the influence of heat, as indicated by the '\textbackslash(Delta)' sign in the reactions given.

In the textbook exercise, we also have replacement reactions as shown in step 8:\[VCl_3(g) + 2K(l) \rightarrow VKCl_2(s) + KCl(s)\]Here, potassium (K) replaces a part of the VCl3 to form new compounds. It is essential to recognize the type of reaction to predict the products and then balance the equation accordingly. Understanding the specific characteristics of reactants helps us anticipate reaction outcomes, which is a key skill in chemistry.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where a single compound breaks down into two or more products. This process often requires an input of energy in the form of heat, light, or electricity. In the classroom, decomposition reactions are often demonstrated using heat, as seen in the given textbook examples, particularly step 4:\[CoCO_3(s) \rightarrow CoO(s) + CO_2(g)\]Under the influence of heat (indicated by the symbol '\textbackslash(Delta)'), the cobalt(II) carbonate decomposes into cobalt(II) oxide and carbon dioxide gas. When balancing decomposition reactions, it's important to pay attention to the number of atoms of each element to ensure the law of conservation of mass is upheld. Meanwhile, identifying the required conditions, such as the heat mentioned above, is also part of understanding and successfully predicting decomposition reactions.