Question

Balance the following equations: (a) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\operatorname{LiI}(a q) \longrightarrow_{\mathrm{LiNO}_{3}}(a q)+\mathrm{HgI}_{2}(s)\) (b) \(\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{ICl}_{5}(s)\) (c) \(\mathrm{Al}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) (d) \(\mathrm{CuSO}_{4}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow\) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (e) \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \longrightarrow\) \(\mathrm{Mn}_{2} \mathrm{~S}_{3}(s)+\mathrm{NaNO}_{3}(a q)\)

Short answer

Balanced equations: (a) 1,2,2,1; (b) 1,5,2; (c) 4,3,2; (d) 1,2,1,1; (e) 2,3,1,6.

Step-by-step solution

Analyze Equation (a)

The equation is: \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(aq)+\mathrm{LiI}(aq) \rightarrow \mathrm{LiNO}_{3}(aq)+\mathrm{HgI}_{2}(s)\). Identify the elements: Hg, NO₃, Li, I. Ensure atoms for each element are equal on both sides. We see one \(\mathrm{Hg}\), two \(\mathrm{NO}_{3}\), one \(\mathrm{Li}\) and two \(\mathrm{I}\) on the left and right.

Balance Iodine (I) for (a)

We have two iodine atoms (I) from \(\mathrm{LiI}\) to form \(\mathrm{HgI}_{2}(s)\). Therefore, the iodide ions are balanced.

Balance Nitrate (NO₃) for (a)

To balance nitrates, note that \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) provides two NO₃ groups for two \(\mathrm{LiNO}_{3}\) molecules. Add coefficient 2 for \(\mathrm{LiNO}_{3}\) and \(\mathrm{LiI}\) to make NO₃ ions balanced: \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(aq) + 2\mathrm{LiI}(aq) \rightarrow 2\mathrm{LiNO}_{3}(aq) + \mathrm{HgI}_{2}(s)\).

Analyze Equation (b)

Equation is: \(\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ICl}_{5}(s)\). Iodine and chlorine need to yield \(\mathrm{ICl}_{5}\). Thus, 5 \(\mathrm{Cl}_{2}\) molecules are required per 2 \(\mathrm{I}_{2}\) to match \(\mathrm{ICl}_{5}\). Use coefficients to balance: \(\mathrm{I}_{2} (s) + 5\mathrm{Cl}_{2} (g) \rightarrow 2\mathrm{ICl}_{5} (s)\).

Analyze Equation (c)

Equation is: \(\mathrm{Al}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s)\). Each \(\mathrm{Al}_{2}\mathrm{O}_{3}\) molecule has 2 Al and 3 O. We need to balance both.

Balance Aluminum (Al) and Oxygen (O) for (c)

Double the \(\mathrm{Al}_{2}\mathrm{O}_{3}\) coefficients, making it 2, leading to 4 Al and 3 O₂ needed. Thus, use 4 for \(\mathrm{Al}\) and 3 for \(\mathrm{O}_{2}\): \(4\mathrm{Al}(s) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Al}_{2}\mathrm{O}_{3}(s)\).

Analyze Equation (d)

Equation is: \(\mathrm{CuSO}_{4}(aq) + \mathrm{AgNO}_{3}(aq) \rightarrow \mathrm{Ag}_{2}\mathrm{SO}_{4}(s) + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq)\). Balance \(\mathrm{Ag}\) and groups SO₄ and NO₃.

Balance Silver (Ag) and Nitrate for (d)

Two \(\mathrm{AgNO}_{3}\) balance \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\), thus, the coefficients for \(\mathrm{AgNO}_{3}\) and resulting \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) are 2. Equation balanced: \(\mathrm{CuSO}_{4}(aq) + 2\mathrm{AgNO}_{3}(aq) \rightarrow \mathrm{Ag}_{2}\mathrm{SO}_{4}(s) + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq)\).

Analyze Equation (e)

Equation is: \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{3}(aq)+\mathrm{Na}_{2}\mathrm{~S}(aq) \rightarrow \mathrm{Mn}_{2}\mathrm{~S}_{3}(s)+\mathrm{NaNO}_{3}(aq)\). Balance Mn and Na.

Balance the Atoms for (e)

For \(\mathrm{Mn}\), multiply \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{3}\) by 2, yielding \(\mathrm{Mn}_{2}\mathrm{~S}_{3}\) and making 6 NO₃. With 3 \(\mathrm{NaNO}_{3}\) formed, using 3 of \(\mathrm{Na}_{2}\mathrm{~S}\), balance: \(2\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{3}(aq) + 3\mathrm{Na}_{2}\mathrm{~S}(aq) \rightarrow \mathrm{Mn}_{2}\mathrm{~S}_{3}(s) + 6\mathrm{NaNO}_{3}(aq)\).

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. When balancing chemical equations, stoichiometry ensures that the law of conservation of mass is adhered to, meaning that the same amount of each element is present before and after the reaction.

In practice, this involves adjusting the coefficients of the compounds involved in a reaction so that the number of atoms for each element matches on both sides of the equation. For example, in the reaction \[\mathrm{Al}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s),\]stoichiometry dictates that we use 4 aluminum atoms and 3 oxygen molecules to yield 2 molecules of aluminum oxide, symbolized by adjusting coefficients: \[4\mathrm{Al}(s) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Al}_{2}\mathrm{O}_{3}(s).\]By understanding and applying stoichiometry, chemists can predict the quantities of products formed in a reaction and the amount of reactants needed. This calculation is crucial in industrial processes where efficiency and cost-effectiveness are paramount.

Chemical Reaction

Chemical reactions are processes where substances, known as reactants, transform into different substances, known as products. Understanding chemical reactions involves recognizing the change occurring at the molecular level and predicting the products formed.

For example, in the chemical reaction: \[\mathrm{I}_{2}(s) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{ICl}_{5}(s),\]iodine and chlorine react to form iodine pentachloride. This transformation requires balancing the reaction to ensure that stoichiometry is satisfied by using appropriate coefficients. Recognizing the intended product and how atoms rearrange helps in writing and balancing chemical reactions effectively.

• Reactants: Substances initially present in a reaction.
• Products: New substances formed as a result of the reaction.
• Direction: Shown via the arrow, indicating the flow from reactants to products.

Understanding these core aspects of a chemical reaction not only helps in balancing but also aids in a deeper appreciation of how substances interact at a molecular level.

Molecular Coefficients

Molecular coefficients are numbers placed in front of compounds in a chemical reaction to balance the equation. They represent the number of moles of a substance involved in the reaction and are essential for maintaining the conservation of mass.

In balancing the chemical equation: \[\mathrm{CuSO}_{4}(aq) + \mathrm{AgNO}_{3}(aq) \rightarrow \mathrm{Ag}_{2}\mathrm{SO}_{4}(s) + \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq),\]coefficients such as 2 for \(\mathrm{AgNO}_{3}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) ensure there is an equal number of atoms for each element on both sides.

Key Benefits of Using Molecular Coefficients:

• Ensure the law of conservation of mass is followed.
• Provide a clear stoichiometric relationship between reactants and products.
• Aid in calculating the amount of each substance needed or produced in a chemical reaction.

Thus, understanding molecular coefficients is crucial for correctly formulating and balancing chemical equations, a skill necessary for both academic and practical applications in the field of chemistry.