Question

Balance the following equations: (a) \(\mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)\) (b) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)

Short answer

The balanced equations are as follows: (a) \(\mathrm{Al}_{4} \mathrm{C}_{3}(s) + 6\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4\mathrm{Al}(\mathrm{OH})_{3}(s) + 3\mathrm{CH}_{4}(g)\) (b) \(2\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l) + 7\mathrm{O}_{2}(g) \longrightarrow 10\mathrm{CO}_{2}(g) + 10\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2\mathrm{Fe}(\mathrm{OH})_{3}(s) + 6\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + 6\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Mg}_{3} \mathrm{N}_{2}(s) + 6\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow 3\mathrm{MgSO}_{4}(a q) + 3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)

Step-by-step solution

Identify the elements

Identify the elements present in the equation: Al, C, H, and O.

Count the atoms

Count the number of atoms for each element: - Reactants: 4 Al, 3 C, 2 H, and 1 O. - Products: 1 Al, 1 C, 7 H, and 4 O.

Adjust the coefficients

Adjust the coefficients to balance the atoms: - To balance Al, add a coefficient of 4 in front of Al(OH)3: \(\mathrm{Al}_{4} \mathrm{C}_{3} + \mathrm{H}_{2} \mathrm{O} \longrightarrow 4\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{CH}_{4}\) - To balance C, add a coefficient of 3 in front of CH4: \(\mathrm{Al}_{4} \mathrm{C}_{3} + \mathrm{H}_{2} \mathrm{O} \longrightarrow 4\mathrm{Al}(\mathrm{OH})_{3}+3\mathrm{CH}_{4}\) - To balance H, add a coefficient of 6 in front of H2O: \(\mathrm{Al}_{4} \mathrm{C}_{3} + 6\mathrm{H}_{2} \mathrm{O} \longrightarrow 4\mathrm{Al}(\mathrm{OH})_{3}+3\mathrm{CH}_{4}\)

Verify the equation is balanced

After adjusting the coefficients, the equation is now balanced as shown below: \[\mathrm{Al}_{4} \mathrm{C}_{3}(s) + 6\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4\mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{CH}_{4}(g)\] #(b) Balancing C5H10O2(l) + O2(g) -> CO2(g) + H2O(g)#

Identify the elements

Identify the elements present in the equation: C, H, and O.

Count the atoms

Count the number of atoms for each element: - Reactants: 5 C, 10 H, and 4 O. - Products: 1 C, 2 H, and 3 O.

Adjust the coefficients

Adjust the coefficients to balance the atoms: - To balance C, add a coefficient of 5 in front of CO2: \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2} + \mathrm{O}_{2} \longrightarrow 5\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) - To balance H, add a coefficient of 5 in front of H2O: \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2} + \mathrm{O}_{2} \longrightarrow 5\mathrm{CO}_{2}+5\mathrm{H}_{2} \mathrm{O}\) - To balance O, add a coefficient of 7/2 in front of O2. Since we need whole-number coefficients, we can multiply the entire equation by 2: \(2\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2} + 7\mathrm{O}_{2} \longrightarrow 10\mathrm{CO}_{2}+10\mathrm{H}_{2} \mathrm{O}\)

Verify the equation is balanced

After adjusting the coefficients, the equation is now balanced as shown below: \[2\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l) + 7\mathrm{O}_{2}(g) \longrightarrow 10\mathrm{CO}_{2}(g)+10\mathrm{H}_{2} \mathrm{O}(g)\] #(c) Balancing Fe(OH)3(s) + H2SO4(aq) -> Fe2(SO4)3(aq) + H2O(l)#

Identify the elements

Identify the elements present in the equation: Fe, O, H, and S.

Count the atoms

Count the number of atoms for each element: - Reactants: 1 Fe, 3 O, 3 H, and 1 S. - Products: 2 Fe, 12 O, 2 H, and 3 S.

Adjust the coefficients

Adjust the coefficients to balance the atoms: - To balance Fe, add a coefficient of 2 in front of Fe(OH)3: \(2\mathrm{Fe}(\mathrm{OH})_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O}\) - To balance O, add a coefficient of 6 in front of H2O: \(2\mathrm{Fe}(\mathrm{OH})_{3} + \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+6\mathrm{H}_{2} \mathrm{O}\) - To balance H, add a coefficient of 6 in front of H2SO4: \(2\mathrm{Fe}(\mathrm{OH})_{3} + 6\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+6\mathrm{H}_{2} \mathrm{O}\) - To balance S, the coefficients are already correct.

Verify the equation is balanced

After adjusting the coefficients, the equation is now balanced as shown below: \[2\mathrm{Fe}(\mathrm{OH})_{3}(s) + 6\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6\mathrm{H}_{2} \mathrm{O}(l)\] #(d) Balancing Mg3N2(s) + H2SO4(aq) -> MgSO4(aq) + (NH4)2SO4(aq)#

Identify the elements

Identify the elements present in the equation: Mg, N, H, S, and O.

Count the atoms

Count the number of atoms for each element: - Reactants: 3 Mg, 2 N, 2 H, 1 S, and 4 O. - Products: 1 Mg, 2 N, 8 H, 2 S, and 8 O.

Adjust the coefficients

Adjust the coefficients to balance the atoms: - To balance Mg, add a coefficient of 3 in front of MgSO4: \(\mathrm{Mg}_{3} \mathrm{N}_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3\mathrm{MgSO}_{4}+\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\) - To balance N, the coefficients are already correct. - To balance H, add a coefficient of 6 in front of H2SO4: \(\mathrm{Mg}_{3} \mathrm{N}_{2} + 6\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3\mathrm{MgSO}_{4}+\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\) - To balance S, add a coefficient of 3 in front of (NH4)2SO4: \(\mathrm{Mg}_{3} \mathrm{N}_{2} + 6\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3\mathrm{MgSO}_{4}+3\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\) - To balance O, the coefficients are already correct.

Verify the equation is balanced

After adjusting the coefficients, the equation is now balanced as shown below: \[\mathrm{Mg}_{3} \mathrm{N}_{2}(s)+6\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow 3\mathrm{MgSO}_{4}(a q)+3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\]

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with determining the quantities of reactants and products involved in a chemical reaction. It's based on the principle that matter is not created or destroyed in a reaction, which is known as the law of conservation of mass. This principle implies that for a chemical equation to represent a real reaction, both sides must have the same number of atoms for each element.

Understanding stoichiometry is essential in balancing chemical equations, a process that ensures the number of each type of atom is equal on both sides of the reaction. For example, in the provided exercise for balancing the reaction between aluminum carbide and water, we must ensure that the aluminum (Al), carbon (C), hydrogen (H), and oxygen (O) atoms are balanced. Through stoichiometric calculations, we can verify that the balanced equation accurately represents the mass and substance relationships within the reaction.

Accurate stoichiometry is not just academic; it's critical in real-world applications such as pharmaceuticals, where precise dosages of reactants are essential, and in industrial processes that must maximize efficiency and minimize waste.

Chemical Reaction

A chemical reaction is a process that involves the transformation of one set of chemical substances into another. This is represented by a chemical equation that shows the reactants on the left side and the products on the right side. Each chemical reaction involves changes in the bonds that join atoms together, which results in the formation of new substances.

When writing or balancing a chemical equation, like those in the exercise, it's crucial to understand that during these reactions, the type of atom does not change, but the way atoms are arranged does. The reactants undergo a transformation, breaking original bonds and forming new ones, which creates the products of the reaction. Each type of chemical reaction has its unique characteristics, which influence how stoichiometry is applied to balance the equation. For example, combustion reactions will always produce carbon dioxide and water, given enough oxygen, while synthesis reactions will create a more complex product from simpler reactants.

Law of Conservation of Mass

The law of conservation of mass is a fundamental concept in chemistry stating that mass cannot be created nor destroyed in a chemical reaction. This means that the mass of the reactants must equal the mass of the products. During a reaction, atoms are neither created nor destroyed; they are simply rearranged to form new substances.

This law underpins the process of balancing chemical equations. For instance, when balancing the equation for the reaction of iron(III) hydroxide with sulfuric acid, we ensure that the number of iron, oxygen, hydrogen, and sulfur atoms in the reactants is equal to the number in the products. If the initial balancing attempt doesn't fulfill the law of conservation of mass, adjustments to coefficients are made until the equation is balanced. These coefficients represent the relative amounts of each substance involved in the reaction and ensure that the mass remains consistent throughout the process.