Question

Balance each of the following oxidation-reduction reactions by inspection. a. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) c. \(\operatorname{SnO}_{2}(s)+C(s) \rightarrow \operatorname{Sn}(s)+\operatorname{CO}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

The balanced equations for the given reactions are: a. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{CO}(g)+2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) c. \(\operatorname{SnO}_{2}(s)+C(s) \rightarrow \operatorname{Sn}(s)+2\operatorname{CO}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\)

Step-by-step solution

Count atoms

On the left side, there are 3 C atoms, 8 H atoms, and 2 O atoms. On the right side, there is 1 C atom, 2 H atoms, and 3 O atoms.

Adjust coefficients

To balance the carbon atoms, we can multiply the CO2 by 3. To balance the hydrogen atoms, we can multiply the H2O by 4. To balance the oxygen atoms, we can multiply the O2 on the left side by 5. The balanced equation for reaction (a) is: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g)+4\mathrm{H}_{2} \mathrm{O}(g)\). For reaction (b):

Count atoms

On the left side, there is 1 C atom, 2 H atoms, and 1 O atom. On the right side, there is 1 C Atom, 4 H atoms, and 1 O atom.

Adjust coefficients

To balance the hydrogen atoms, we can multiply the H2 on the left side by 2. The balanced equation for reaction (b) is: \(\mathrm{CO}(g)+2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\). For reaction (c):

Count atoms

On the left side, there is 1 Sn atom, 2 O atoms, and 1 C atom. On the right side, there is 1 Sn atom, 1 C atom, and 1 O atom.

Adjust coefficients

To balance the oxygen atoms, we can multiply the CO on the right side by 2. The balanced equation for reaction (c) is: \(\operatorname{SnO}_{2}(s)+C(s) \rightarrow \operatorname{Sn}(s)+2\operatorname{CO}(g)\). For reaction (d):

Count atoms

On the left side, there are 2 C atoms, 6 H atoms, and 3 O atoms. On the right side, there is 1 C atom, 2 H atoms, and 3 O atoms.

Adjust coefficients

To balance the carbon atoms, we can multiply the CO2 on the right side by 2. To balance the hydrogen atoms, we can multiply the H2O on the right side by 3. To balance the oxygen atoms, we can multiply the O2 on the left side by 1.5. The balanced equation for reaction (d) is: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\frac{3}{2}\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\).

Key concepts

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is upheld: the number of atoms of each element must be the same on both sides of the reaction. It involves adjusting the coefficients (the numbers before chemical formulas) to get the same number of atoms for each element in both reactants and products.

For example, in the oxidation-reduction reaction \texttt{\(\mathrm{C}_3 \mathrm{H}_8(g)+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(g)\)},initially, there are imbalances in the number of carbon (C), hydrogen (H), and oxygen (O) atoms. The balancing process corrects this by making sure that there is an equal amount of each element on both sides: producing \texttt{\(\mathrm{C}_3 \mathrm{H}_8(g)+5\mathrm{O}_2(g) \rightarrow 3\mathrm{CO}_2(g)+4\mathrm{H}_2 \mathrm{O}(g)\).
It's important to start with substances that contain elements that appear only once on each side, and to leave more complex molecules and polyatomic ions (if unaltered during the reaction) for last.

Stoichiometry

Stoichiometry, at its core, involves the calculation of relative quantities of reactants and products in chemical reactions. It is founded on the principle of balanced equations; hence, before you can dive into stoichiometric calculations, ensuring the chemical equation is balanced is crucial.

Stoichiometry allows us to predict how much product is formed from a certain amount of reactants or how much of the reactants are necessary to create a certain amount of product. For instance, in the reaction \texttt{\(\mathrm{CO}(g)+2\mathrm{H}_2(g) \rightarrow \mathrm{CH}_3 \mathrm{OH}(l)\)},if we have 3 moles of \(\mathrm{H}_2\), stoichiometry tells us we would produce 1.5 moles of \(\mathrm{CH}_3 \mathrm{OH}\) because the stoichiometric coefficients indicate two moles of hydrogen gas are needed for each mole of methanol produced.

Understanding stoichiometry is essential for everything from cooking recipes to industrial chemical manufacturing, as it ensures the correct proportions are used to achieve desired outcomes without wasting resources.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are processes where the oxidation state of atoms or molecules change due to the transfer of electrons. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.

In a redox reaction, such as \texttt{\(\operatorname{SnO}_2(s)+C(s) \rightarrow \operatorname{Sn}(s)+2\operatorname{CO}(g)\)},\(\operatorname{SnO}_2\) is reduced to \(\operatorname{Sn}\) as it gains electrons, and carbon (C) is oxidized to carbon monoxide (CO) as it loses electrons. Every redox reaction consists of two parts, the oxidation and the reduction, which occur simultaneously—a concept known as the redox pair.

To identify redox reactions, look for changes in oxidation number and then balance the electrons gained and lost. Also, remember that the total number of electrons lost in oxidation must equal the total number gained in reduction. In complex reactions, using the half-reaction method can simplify the process of balancing redox reactions by treating the oxidation and reduction processes separately before combining them. This approach is particularly effective in reactions where multiple species undergo redox changes.