Question

Balance each of the following chemical equations by inspection: (a) \(\mathrm{Co}(\mathrm{s})+\mathrm{O}_{2}(g) \rightarrow \mathrm{Co}_{2} \mathrm{O}_{3}(\mathrm{~s})\) (b) \(\mathrm{LiClO}_{3}(s) \rightarrow \mathrm{LiCl}(s)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{Cu}(\mathrm{s})+\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{Ag}(s)\) (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{LiCl}(a q) \rightarrow \mathrm{PbCl}_{2}(\mathrm{~s})+\mathrm{LiNO}_{3}(a q)\) (e) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

(a) \( 4\mathrm{Co} + 3\mathrm{O}_{2} \rightarrow 2\mathrm{Co}_{2}\mathrm{O}_{3} \) (b) \( 2\mathrm{LiClO}_{3} \rightarrow 2\mathrm{LiCl} + 3\mathrm{O}_{2} \) (c) \( \mathrm{Cu} + 2\mathrm{AgC}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \rightarrow \mathrm{Cu}(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2})_{2} + 2\mathrm{Ag} \) (d) \( \mathrm{Pb}(\\mathrm{NO}_{3})_{2} + 2\mathrm{LiCl} \rightarrow \mathrm{PbCl}_{2} + 2\mathrm{LiNO}_{3} \) (e) \( 3\mathrm{H}_{2}\mathrm{SO}_{4} + 2\mathrm{Al}(\mathrm{OH})_{3} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} + 6\mathrm{H}_{2}\mathrm{O} \)

Step-by-step solution

Balancing Equation (a)

The unbalanced equation is \( \mathrm{Co}(\mathrm{s}) + \mathrm{O}_{2}(g) \rightarrow \mathrm{Co}_{2} \mathrm{O}_{3}(\mathrm{s}) \). 1. Count the number of each atom on both sides. Initially, Co=1, O=2 on the left and Co=2, O=3 on the right. 2. Balance Co by placing a coefficient of 2 in front of \( \mathrm{Co} \). Now, Co=2 on both sides, but O still needs balancing. 3. Adjust the O atoms by ensuring they add up to 3 on both sides. Place a coefficient of \( \frac{3}{2} \) in front of \( \mathrm{O}_{2} \) so that O=3. 4. To deal with fractional coefficients, multiply the entire equation by 2: \[ 4 \mathrm{Co}(\mathrm{s}) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Co}_{2} \mathrm{O}_{3}(\mathrm{s}) \]. Now the equation is balanced.

Balancing Equation (b)

The unbalanced equation is \( \mathrm{LiClO}_{3}(s) \rightarrow \mathrm{LiCl}(s) + \mathrm{O}_{2}(g) \). 1. Count atoms: Initially, Li=1, Cl=1, O=3 on the left, and Li=1, Cl=1, O=2 on the right. 2. Notice that O atoms are unbalanced: 3 on the left, 2 on the right. To balance O atoms, place \( \frac{3}{2} \) in front of \( \mathrm{O}_{2} \). 3. To remove the fraction, multiply the whole equation by 2: \[ 2 \mathrm{LiClO}_{3}(s) \rightarrow 2 \mathrm{LiCl}(s) + 3 \mathrm{O}_{2}(g) \]. Now the equation is balanced.

Balancing Equation (c)

The unbalanced equation is \( \mathrm{Cu}(\mathrm{s}) + \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q) + \mathrm{Ag}(s) \).1. Count atoms: Initially, Cu=1, Ag=1, C=2, H=3, O=2 on the left, Cu=1, Ag=1, C=4, H=6, O=4 on the right. 2. Balance Ag by placing a coefficient of 2 in front of \( \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \). Now Ag=2, C=4, H=6, O=4 on both sides. 3. Adjust Cu by placing a coefficient of 2 in front of \( \mathrm{Ag} \). Now both sides are balanced: Cu=1, Ag=2, C=4, H=6, O=4. 4. Final balanced equation: \( \mathrm{Cu}(\mathrm{s}) + 2 \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q) + 2 \mathrm{Ag}(s) \).

Balancing Equation (d)

The unbalanced equation is \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) + \mathrm{LiCl}(a q) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) + \mathrm{LiNO}_{3}(a q) \). 1. Start by counting the atoms: Pb=1, N=2, O=6, Li=1, Cl=1 on the left, and Pb=1, Cl=2, Li=1, N=1, O=3 on the right. 2. Balance Cl atoms by placing a 2 in front of \( \mathrm{LiCl} \). Now Cl atoms are balanced (Cl=2), but Li and N need adjusting. 3. To balance N atoms, place a 2 in front of \( \mathrm{LiNO}_{3} \). 4. Final balanced equation: \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) + 2 \mathrm{LiCl}(a q) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) + 2 \mathrm{LiNO}_{3}(a q) \).

Balancing Equation (e)

The unbalanced equation is \( \mathrm{H}_{2} \mathrm{SO}_{4}(a q) + \mathrm{Al}(\mathrm{OH})_{3}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + \mathrm{H}_{2} \mathrm{O}(l) \). 1. Count atoms: H=2, S=1, O=7, Al=1 on the left; Al=2, S=3, O=13, H=2 on the right.2. Balance Al atoms by placing a 2 in front of \( \mathrm{Al}(\mathrm{OH})_{3} \). Now Al=2, H=8, and O=13 on the left.3. Balance S atoms by placing a 3 in front of \( \mathrm{H}_{2} \mathrm{SO}_{4} \). Now S and Al are balanced.4. Balance H and O by placing a 6 in front of \( \mathrm{H}_{2} \mathrm{O} \). This makes: H=12 and O=13 on both sides. 5. Final balanced equation: \( 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) + 2 \mathrm{Al}(\mathrm{OH})_{3}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + 6 \mathrm{H}_{2} \mathrm{O}(l) \).

Key concepts

Stoichiometry

Stoichiometry is a fascinating concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is crucial for balancing chemical equations and involves understanding the molar relationships between the substances involved.
The steps include determining the amount of one substance when given the amount of another. This comes in handy when you need to know how much reactant is needed or how much product will be formed in a reaction.

• The coefficients in a balanced equation are used as the conversion factors for these calculations.
• These coefficients represent the number of moles of each compound that participate in the reaction.

Therefore, stoichiometry is all about understanding the quantity relationships within a chemical reaction. Each balanced equation can be used to derive these relationships, which are utterly important in quantitative chemistry. Understanding this equips one with the ability to predict the outcomes and necessities of chemical processes.

Chemical Reactions

Chemical reactions are processes that lead to the transformation of one set of chemical substances to another. They are characterized by a chemical change, and they are described by chemical equations, which symbolically depict the reactants and the products.
In any chemical reaction, the starting substances are called reactants, and they undergo a transformation to produce the resulting substances known as products.

• For example, when balancing an equation like \[\mathrm{Co} + \mathrm{O}_{2} \rightarrow \mathrm{Co}_{2} \mathrm{O}_{3}\], Co and \[\mathrm{O}_{2}\] are reactants, while \[\mathrm{Co}_{2} \mathrm{O}_{3}\] is the product.
• Reactants must be correctly represented to ensure that the chemical equation is balanced and accurately shows the transformation process.

Balancing a chemical equation ensures that it follows the laws of chemistry, specifically, the law of conservation of mass. This is crucial as it maintains the proportion of each element on either side of the equation, ensuring the reaction is realistic and follows natural laws.

Atom Conservation

Atom conservation is a fundamental principle in chemistry that states atoms are not created or destroyed in chemical reactions. Instead, they are rearranged to form new substances. This principle is pivotal in balancing chemical equations.
When balancing chemical equations, one must ensure that the number of each type of atom on the reactant side of the equation is equal to the number on the product side.

• For instance, if you have \[\mathrm{LiClO}_{3} \rightarrow \mathrm{LiCl} + \mathrm{O}_{2}\], the number of Li, Cl, and O atoms on the left must match those on the right.
• This may involve placing suitable coefficients before compounds in the reaction to equalize the number of atoms.

This balancing acts as a check, confirming that the reaction adheres to the atom conservation law. By ensuring that chemical equations reflect this conservation, scientists and students alike can model real-world chemical processes accurately.