Question

Complete and balance the following equations:$$\begin{array}{l}{\text { (a) } \mathrm{Mg}_{3} \mathrm{N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ {\text { (b) } \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow} \\ {\text { (c) } \mathrm{MnO}_{2}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}}\end{array}$$ $$\begin{array}{l}{\text { (d) } \operatorname{AlP}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ {\text { (e) } \mathrm{Na}_{2} \mathrm{S}(s)+\mathrm{HCl}(a q) \longrightarrow}\end{array}$$

Short answer

The balanced equations are: a) \(3Mg_3N_2(s) + 6H_2O(l) \rightarrow 6MgO(s) + 6NH_3(g)\) b) \(C_3H_7OH(l) + \dfrac{9}{2} O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\) c) \(MnO_2(s) + C(s) \stackrel{\Delta}{\longrightarrow} Mn(s) + CO(g)\) d) \(AlP(s) + 3H_2O(l) \rightarrow Al(OH)_3(s) + PH_3(g)\) e) \(Na_2S(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2S(g)\)

Step-by-step solution

a) Balancing Equation (a)

For equation (a), we have: Mg₃N₂(s) + H₂O(l) → ? We notice that magnesium forms an oxide (MgO) and nitrogen forms an ammonia molecule (NH₃). So, let's rewrite the equation: Mg₃N₂(s) + H₂O(l) → MgO + NH₃ Now, balance the equation: 3Mg₃N₂(s) + 6H₂O(l) → 6MgO + 6NH₃ The balanced equation is given by: \(3Mg_3N_2(s) + 6H_2O(l) \rightarrow 6MgO(s) + 6NH_3(g)\)

b) Balancing Equation (b)

For equation (b), we have: C₃H₇OH (l) + O₂(g) → ? We notice that this is a combustion reaction, producing carbon dioxide (CO₂) and water (H₂O): C₃H₇OH (l) + O₂(g) → CO₂(g) + H₂O(l) Now, balance the equation: C₃H₇OH (l) + 9/2 O₂(g) → 3CO₂(g) + 4H₂O(l) The balanced equation is given by: \(C_3H_7OH(l) + \dfrac{9}{2} O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\)

c) Balancing Equation (c)

For equation (c), we have: MnO₂(s) + C(s) → ? We notice that this is a redox reaction, where MnO₂ is reduced to Mn and carbon is oxidized to CO: MnO₂(s) + C(s) → Mn(s) + CO(g) Now, balance the equation: MnO₂(s) + C(s) → Mn(s) + CO(g) The balanced equation is already given by: \(MnO_2(s) + C(s) \stackrel{\Delta}{\longrightarrow} Mn(s) + CO(g)\)

d) Balancing Equation (d)

For equation (d), we have: AlP (s) + H₂O (l) → ? AlP reacts with water to produce a metal hydroxide (Al(OH)₃) and a phosphine gas (PH₃): AlP (s) + H₂O (l) → Al(OH)₃(s) + PH₃(g) Now, balance the equation: AlP (s) + 3H₂O (l) → Al(OH)₃(s) + PH₃(g) The balanced equation is given by: \(AlP(s) + 3H_2O(l) \rightarrow Al(OH)_3(s) + PH_3(g)\)

e) Balancing Equation (e)

For equation (e), we have: Na₂S(s) + HCl(aq) → ? Na₂S reacts with HCl to produce a metal chloride (NaCl) and a hydrogen sulfide gas (H₂S): Na₂S(s) + HCl(aq) → NaCl(aq) + H₂S(g) Now, balance the equation: Na₂S(s) + 2HCl(aq) → 2NaCl(aq) + H₂S(g) The balanced equation is given by: \(Na_2S(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2S(g)\)

Key concepts

Stoichiometry

Stoichiometry is like the recipe for chemistry. It involves using the coefficients in a balanced chemical equation to calculate the relative amounts of reactants and products involved in a reaction.

Imagine baking cookies—you need the right proportions of sugar, flour, and butter. In chemistry, the balanced equation tells you the proportions of chemicals that react together. For instance, in equation (a) from our exercise, the balanced reaction indicates that 3 molecules of Mg₃N₂ will react with 6 molecules of H₂O to produce 6 molecules of MgO and 6 molecules of NH₃.

To balance a chemical equation, we need to make sure that the number of atoms of each element is the same on both sides. This reflects the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

Redox Reactions

Redox reactions are all about the transfer of electrons between substances. One substance will give up electrons (oxidation) while another will gain electrons (reduction).

For instance, take the reaction of MnO₂ with carbon (equation c). In this reaction, MnO₂ is reduced to Mn because it gains electrons, and carbon is oxidized to CO because it loses electrons. Here's an easy way to remember: OIL RIG—Oxidation Is Loss, Reduction Is Gain.

To identify a redox reaction, look for changes in the oxidation states of the elements. In a balanced redox equation, the total increase in oxidation state must be balanced by a total decrease in oxidation state, because the number of electrons lost in oxidation must equal the number of electrons gained in reduction.

Combustion Reaction

A combustion reaction is basically a fire in a chemical equation. It usually involves a substance (usually a hydrocarbon) reacting with oxygen to release energy in the form of heat and light.

Equation (b) is a classic combustion reaction where the hydrocarbon (C₃H₇OH) combusts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). This type of reaction always needs oxygen and will produce CO₂ and H₂O if a hydrocarbon is the fuel.

To balance a combustion reaction, start by balancing the carbons and hydrogens and end with the oxygens, as the O₂ molecule can be adjusted to balance the reaction, like splitting the O₂ molecule into fractions as shown in the exercise. Combustion reactions are exothermic, meaning they release energy, which is why they are often associated with fire and burning.