Problem 67 - Chemistry and Chemical Reactivity

Question

Complete and balance the following equations. (a) \(\mathrm{KClO}_{3}+\) heat \(\rightarrow\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow\) (c) \(\mathrm{Na}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow\) (d) \(\mathrm{P}_{4}(\mathrm{s})+\mathrm{KOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) (e) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})+\) heat \(\rightarrow\) (f) \(\operatorname{In}(\mathrm{s})+\mathrm{Br}_{2}(\ell) \rightarrow\) (g) \(\mathrm{SnCl}_{4}(\ell)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\)

Step-by-Step Solution

Answer

(a) \(2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2\); (b) \(2\mathrm{H}_2\mathrm{S} + 3\mathrm{O}_2 \rightarrow 2\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}\); (c) \(4\mathrm{Na} + \mathrm{O}_2 \rightarrow 2\mathrm{Na}_2\mathrm{O}\); (d) \(\mathrm{P}_4 + 3\mathrm{KOH} + 3\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{KH}_2\mathrm{PO}_2 + \mathrm{PH}_3\); (e) \(\mathrm{NH}_4\mathrm{NO}_3 \rightarrow \mathrm{N}_2\mathrm{O} + 2\mathrm{H}_2\mathrm{O}\); (f) \(2\mathrm{In} + 3\mathrm{Br}_2 \rightarrow 2\mathrm{InBr}_3\); (g) \(\mathrm{SnCl}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SnO}_2 + 4\mathrm{HCl}\).

Step 1: Decompose Potassium Chlorate

The decomposition of potassium chlorate \(\mathrm{KClO}_3\) upon heating produces potassium chloride and oxygen gas. The balanced equation is:\[2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2\]

Step 2: Combustion of Hydrogen Sulfide

When hydrogen sulfide \(\mathrm{H}_2\mathrm{S}\) reacts with oxygen gas \(\mathrm{O}_2\), it produces sulfur dioxide \(\mathrm{SO}_2\) and water. The balanced equation is:\[2\mathrm{H}_2\mathrm{S} + 3\mathrm{O}_2 \rightarrow 2\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}\]

Step 3: Sodium and Oxygen Reaction

Sodium \(\mathrm{Na}\) reacts with oxygen \(\mathrm{O}_2\) to form sodium oxide \(\mathrm{Na}_2\mathrm{O}\). The balanced equation is:\[4\mathrm{Na} + \mathrm{O}_2 \rightarrow 2\mathrm{Na}_2\mathrm{O}\]

Step 4: Reaction of Phosphorus with KOH and Water

Phosphorus \(\mathrm{P}_4\) reacts with potassium hydroxide \(\mathrm{KOH}\) and water \(\mathrm{H}_2\mathrm{O}\) to produce phosphine \(\mathrm{PH}_3\) and potassium hypophosphite \(\mathrm{KH}_2\mathrm{PO}_2\). The balanced equation is:\[\mathrm{P}_4 + 3\mathrm{KOH} + 3\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{KH}_2\mathrm{PO}_2 + \mathrm{PH}_3\]

Step 5: Decomposition of Ammonium Nitrate

Ammonium nitrate \(\mathrm{NH}_4\mathrm{NO}_3\) decomposes upon heating to produce nitrous oxide \(\mathrm{N}_2\mathrm{O}\) and water. The balanced equation is:\[\mathrm{NH}_4\mathrm{NO}_3 \rightarrow \mathrm{N}_2\mathrm{O} + 2\mathrm{H}_2\mathrm{O}\]

Step 6: Reaction of Indium with Bromine

Indium \(\mathrm{In}\) reacts with bromine \(\mathrm{Br}_2\) to form indium tribromide \(\mathrm{InBr}_3\). The balanced equation is:\[2\mathrm{In} + 3\mathrm{Br}_2 \rightarrow 2\mathrm{InBr}_3\]

Step 7: Hydrolysis of Tin(IV) Chloride

Tin(IV) chloride \(\mathrm{SnCl}_4\) reacts with water \(\mathrm{H}_2\mathrm{O}\) to form tin dioxide \(\mathrm{SnO}_2\) and hydrochloric acid \(\mathrm{HCl}\). The balanced equation is:\[\mathrm{SnCl}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SnO}_2 + 4\mathrm{HCl}\]

Key Concepts

Decomposition Reactions

Decomposition reactions involve a single compound breaking down into two or more simpler substances. These reactions often require an input of energy, such as heat, light, or electricity, to occur. In the context of the original exercise, potassium chlorate (\[\mathrm{KClO}_3\]) undergoes decomposition when heated. During this process, it breaks down into potassium chloride (\[\mathrm{KCl}\]) and oxygen gas (\[\mathrm{O}_2\]). This can be expressed in the balanced chemical equation:
\[2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2\]
Decomposition reactions are important in various fields, such as in the production of oxygen from substances and understanding chemical stability. Recognizing these reactions helps students predict the outcomes of heating certain compounds.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen and releases energy in the form of heat and light. These reactions are essential in energy production, as they are the basis for many everyday processes such as burning fuels in car engines. In the exercise, hydrogen sulfide (\[\mathrm{H}_2\mathrm{S}\]) undergoes combustion by reacting with oxygen (\[\mathrm{O}_2\]) to form sulfur dioxide (\[\mathrm{SO}_2\]) and water (\[\mathrm{H}_2\mathrm{O}\]). The balanced equation for this reaction is:
\[2\mathrm{H}_2\mathrm{S} + 3\mathrm{O}_2 \rightarrow 2\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}\]
Combustion reactions typically produce heat and can happen rapidly or slowly, influencing everything from industrial processes to how energy efficiently we use fuels.

Synthesis Reactions

Synthesis reactions involve two or more substances combining to form a more complex compound. These reactions are the opposite of decomposition and are fundamental in the creation of new molecules. In the exercise, sodium (\[\mathrm{Na}\]) combines with oxygen (\[\mathrm{O}_2\]) to produce sodium oxide (\[\mathrm{Na}_2\mathrm{O}\]), as shown in the balanced equation:
\[4\mathrm{Na} + \mathrm{O}_2 \rightarrow 2\mathrm{Na}_2\mathrm{O}\]
Synthesis reactions are crucial in manufacturing complex products, such as pharmaceuticals and new materials. Understanding these reactions is important for students because it allows them to see how individual elements can come together to form substances with entirely different properties.

Hydrolysis Reactions

Hydrolysis reactions involve the reaction of a chemical compound with water, leading to the decomposition of that compound. This is vital in organic chemistry and biochemistry, where it aids in breaking down polymers into monomers. A practical example of hydrolysis was given in this exercise with tin(IV) chloride (\[\mathrm{SnCl}_4\]) reacting with water (\[\mathrm{H}_2\mathrm{O}\]). This reaction forms tin dioxide (\[\mathrm{SnO}_2\]) and hydrochloric acid (\[\mathrm{HCl}\]), represented by:
\[\mathrm{SnCl}_4 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{SnO}_2 + 4\mathrm{HCl}\]

• Hydrolysis is key in biological systems for digestion and metabolic processes.
• It helps breakdown complex molecules into usable forms for cells.

Recognizing hydrolysis allows students to understand how water can be a crucial reactant in chemical reactions.