Question

Predict the products and write a balanced molecular equation for each reaction. If no reaction occurs, write NO REACTION. a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow\) b. \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{LiOH}(a q) \longrightarrow\) c. liquid pentanol \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\right)\) and gaseous oxygen d. aqueous strontium sulfide and aqueous copper(II) sulfate

Short answer

a. NO REACTION b. \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(\mathrm{aq}) + 3\mathrm{LiOH}(\mathrm{aq}) \longrightarrow \mathrm{Cr}\left(\mathrm{OH}\right)_{3}(\mathrm{s}) + 3\mathrm{LiNO}_{3}(\mathrm{aq})\) c. \(1\mathrm{C}_{5}\mathrm{H}_{12}\mathrm{O}(l) + 8\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2}\mathrm{O}(l)\) d. \(\mathrm{SrS}(\mathrm{aq}) + \mathrm{CuSO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{SrSO}_{4}(\mathrm{s}) + \mathrm{CuS}(\mathrm{s})\)

Step-by-step solution

Predicting Products for Reaction a

Analyze the reactants. Both \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and \(\mathrm{HNO}_{3}\) are strong acids. Mixing two strong acids does not result in a reaction, thus no new products are formed.

Predicting Products for Reaction b

Recognize the double displacement reaction where the anion from one reactant combines with the cation from the other reactant. The products will be \(\mathrm{Cr}\left(\mathrm{OH}\right)_{3}\) and \(\mathrm{LiNO}_{3}\). Write the unbalanced equation: \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(\mathrm{aq}) + 3\mathrm{LiOH}(\mathrm{aq}) \longrightarrow \mathrm{Cr}\left(\mathrm{OH}\right)_{3}(\mathrm{s}) + 3\mathrm{LiNO}_{3}(\mathrm{aq})\).

Predicting Products for Reaction c

Identify the reaction type as combustion where an organic compound reacts with oxygen to form carbon dioxide and water. Write the unbalanced reaction: \(\mathrm{C}_{5}\mathrm{H}_{12}\mathrm{O}(l) + \mathrm{O}_{2}(g)\longrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\).

Balancing Equation for Reaction c

Balance the reaction coefficients to adhere to the law of conservation of mass. The balanced chemical equation is \(1\mathrm{C}_{5}\mathrm{H}_{12}\mathrm{O}(l) + 8\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2}\mathrm{O}(l)\).

Predicting Products for Reaction d

Consider the double displacement reaction. The products will be strontium sulfate \(\mathrm{SrSO}_{4}\) and copper(II) sulfide \(\mathrm{CuS}\). Write the unbalanced equation: \(\mathrm{SrS}(\mathrm{aq}) + \mathrm{CuSO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{SrSO}_{4}(\mathrm{s}) + \mathrm{CuS}(\mathrm{s})\).

Balancing Equation for Reaction d

Check the stoichiometry of the reaction. The reaction is already balanced, so no further actions are required.

Key concepts

Balancing Chemical Equations

Chemical reactions must be represented by balanced equations to reflect the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Thus, the quantity of each element must remain unchanged before and after the reaction.

To balance a chemical equation, ensure that the number of atoms for each element is equal on both sides. This is achieved by adjusting coefficients, which are the numbers placed before compounds in an equation. For example, in the combustion of pentanol, the balanced equation is \(1\mathrm{C}_5\mathrm{H}_{12}\mathrm{O}(l) + 8\mathrm{O}_2(g) \longrightarrow 5\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(l)\), where the coefficients are adjusted so that the number of carbon, hydrogen, and oxygen atoms on both sides are equal.

Following these steps ensures balanced chemical equations:

• Write the unbalanced equation with the correct formulas for all reactants and products.
• Determine the number of atoms of each element in the reactants and products.
• Adjust coefficients, starting with the most complex molecule, to balance each element one at a time.
• Double check that all elements are balanced and that the coefficients are in the simplest whole-number ratio.

Double Displacement Reaction

Double displacement reactions, also known as metathesis reactions, occur when parts of two ionic compounds are exchanged and form two new compounds. This is characteristic of occurrences in aqueous solution where the anions and cations of the dissolved substances switch partners. The general form is \(AX + BY \longrightarrow AY + BX\).

For example, when aqueous solutions of chromium(III) nitrate and lithium hydroxide are mixed, they undergo a double displacement reaction to form chromium(III) hydroxide and lithium nitrate. This is written as \(\mathrm{Cr}\left(\mathrm{NO}_3\right)_3(\mathrm{aq}) + 3\mathrm{LiOH}(\mathrm{aq}) \longrightarrow \mathrm{Cr}\left(\mathrm{OH}\right)_3(\mathrm{s}) + 3\mathrm{LiNO}_3(\mathrm{aq})\).

Common indicators of a double displacement reaction include the formation of a precipitate, the production of a gas, or a change in color. Predicting the products involves knowledge of solubility rules to determine if any of the products will be insoluble in the reaction mixture.

Combustion Reaction

Combustion reactions are exothermic reactions where a substance (typically a hydrocarbon) reacts with oxygen to release energy in the form of heat and light. The products of complete combustion of hydrocarbons are carbon dioxide and water. This type of reaction is prevalent in everyday life, such as the burning of gasoline in an engine or natural gas in a stove.

An example is the combustion of liquid pentanol, \(\mathrm{C}_5\mathrm{H}_{12}\mathrm{O}\), with gaseous oxygen. The products are carbon dioxide and water, expressed in the balanced combustion reaction: \(1\mathrm{C}_5\mathrm{H}_{12}\mathrm{O}(l) + 8\mathrm{O}_2(g) \longrightarrow 5\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(l)\).

To identify a combustion reaction, look for a reactant that is a hydrocarbon or similar organic molecule, and the presence of oxygen as a reactant. The key to solving these equations lies in balancing the number of carbon, hydrogen, and oxygen atoms to ensure the Law of Conservation of Mass is upheld.