Question

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

Short answer

The short answers for the balanced equations are: (a) \(Mg_3 N_2(s) + 4 H_2 O(l) \to 3 Mg(OH)_2(s) + 2 NH_3(g)\) (b) \(2 C_3 H_7 OH (l) + 9 O_2 (g) \to 6 CO_2 (g) + 8 H_2 O (l)\) (c) \(MnO_2 (s) + C(s) \stackrel{\Delta}{\to} Mn(s) + CO(g)\) (d) \(AlP(s) + 3 H_2 O (l) \to Al(OH)_3 (s) + PH_3 (g)\) (e) \(Na_2 S(s) + 2 HCl (aq) \to 2 NaCl(aq) + H_2 S(aq)\)

Step-by-step solution

(a) Balancing Mg3N2 + H2O

To complete and balance this equation, we need to identify the products formed. Mg3N2(s) + H2O(l) -> Mg(OH)2 (s) + NH3 (g) Now, we will balance the equation: 1. Balance Mg: There are 3 Mg atoms on the left side, so we need 3 Mg(OH)2 on the right side. Mg3N2 + H2O -> 3Mg(OH)2 + NH3 2. Balance N: There are 2 N atoms on the left side, so we need 2 NH3 on the right side. Mg3N2 + H2O -> 3Mg(OH)2 + 2NH3 3. Balance H: We can see there are 2 + 2x3 = 8 H atoms on the right side, so we need 4 H2O on the left side. Mg3N2 + 4H2O -> 3Mg(OH)2 + 2NH3 The balanced equation is: \(Mg_3 N_2(s) + 4 H_2 O(l) \to 3 Mg(OH)_2(s) + 2 NH_3(g)\)

(b) Balancing C3H7OH + O2

This reaction involves the combustion of the alcohol C3H7OH. In combustion reactions, oxygen (O2) reacts with a hydrocarbon to produce carbon dioxide (CO2) and water (H2O). So, the products of this reaction will be CO2 and H2O. C3H7OH(l) + O2(g) -> CO2(g) + H2O(l) Now, let's balance the equation: 1. Balance C: We have 3 C atoms on the left, so we need 3 CO2 molecules on the right. C3H7OH + O2 -> 3CO2 + H2O 2. Balance H: We have 8 H atoms on the left, therefore, we need 4 H2O molecules on the right. C3H7OH + O2 -> 3CO2 + 4H2O 3. Balance O: There are 9 O atoms on the right side, so we need 9/2 O2 molecules on the left side. C3H7OH + 9/2 O2 -> 3CO2 + 4H2O Since fractional coefficients are not allowed in balanced equations, we can multiply the entire equation by 2 to eliminate the fraction: 2C3H7OH(l) + 9O2(g) -> 6CO2(g) + 8H2O(l) The balanced equation is: \(2 C_3 H_7 OH (l) + 9 O_2 (g) \to 6 CO_2 (g) + 8 H_2 O (l)\)

(c) Balancing MnO2 + C

This reaction is a reduction of manganese dioxide (MnO2) by carbon (C) to form elemental manganese (Mn) and carbon monoxide (CO). MnO2(s) + C(s) -> Mn(s) + CO(g) In this case, the equation is already balanced: \(MnO_2 (s) + C(s) \stackrel{\Delta}{\to} Mn(s) + CO(g)\)

(d) Balancing AlP + H2O

This reaction involves the hydrolysis of aluminium phosphide (AlP) in the presence of water (H2O). The products formed are aluminium hydroxide (Al(OH)3) and phosphine (PH3): AlP(s) + H2O(l) -> Al(OH)3(s) + PH3(g) Now, let's balance the equation: 1. Balance Al: There is already one Al atom on each side. 2. Balance P: There is already one P atom on each side. 3. Balance H: There are 6 H atoms on the right side, so we need 3 H2O on the left side. AlP + 3H2O -> Al(OH)3 + PH3 The balanced equation is: \(AlP(s) + 3 H_2 O (l) \to Al(OH)_3 (s) + PH_3 (g)\)

(e) Balancing Na2S + HCl

This is an acid-base reaction, where sodium sulfide (Na2S) reacts with hydrochloric acid (HCl) to produce sodium chloride (NaCl) and hydrogen sulfide (H2S). Na2S(s) + HCl(aq) -> NaCl(aq) + H2S(aq) Now, let's balance the equation: 1. Balance Na: There are 2 Na atoms on the left side, so we need 2 NaCl on the right side. Na2S + HCl -> 2NaCl + H2S 2. Balance S: There is already one S atom on each side. 3. Balance H: There are 4 H atoms on the right side, so we need 2 HCl on the left side. Na2S + 2HCl -> 2NaCl + H2S The balanced equation is: \(Na_2 S(s) + 2 HCl (aq) \to 2 NaCl(aq) + H_2 S(aq)\)

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass where the total mass of reactants equals the total mass of products.

To balance a chemical equation, every element must have the same number of atoms on both the reactant and product sides. Balancing involves adjusting the coefficients (the numbers in front of compounds) to achieve the same number of atoms of each element in the reactants and products. This process requires an understanding of the molecular formulas, as well as a systematic approach to ensure that all atoms are accounted for.

For example, in balancing the reaction \(Mg_3N_2 + 4H_2O \rightarrow 3Mg(OH)_2 + 2NH_3\), we had to adjust the coefficients in front of each compound systematically until the number of atoms for each element was balanced on both sides of the equation.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance (typically a hydrocarbon) combines with oxygen to produce carbon dioxide, water, and energy, usually in the form of heat or light.

The general form of a combustion reaction can be represented as \(\text{fuel} + O_2 \rightarrow CO_2 + H_2O\). In balancing combustion reactions, one usually starts with the carbon and hydrogen atoms before balancing the oxygen atoms, since oxygen is present in the reactants and products. Balancing the equation for the combustion of propanol, \(2C_3H_7OH + 9O_2 \rightarrow 6CO_2 + 8H_2O\), requires a careful approach to ensure that both the reactants and the products have the same number of atoms for each element, with particular attention to the diatomic nature of oxygen gas.

Acid-Base Reactions

Acid-base reactions are fundamental chemical reactions where an acid and a base react to form a salt and usually water. According to the Arrhenius definition, an acid is a substance that increases the concentration of hydrogen ions \(H^+\) in a solution, while a base provides hydroxide ions \(OH^-\).

The general equation for an acid-base reaction is \(\text{acid} + \text{base} \rightarrow \text{salt} + H_2O\). When solving these reactions, stoichiometry plays a vital role in determining the correct proportions of reactants. For instance, balancing the reaction between sodium sulfide and hydrochloric acid, \(Na_2S + 2HCl \rightarrow 2NaCl + H_2S\), reveals the production of sodium chloride (table salt) and hydrogen sulfide as the reaction proceeds.

Reduction Reactions

Reduction reactions, often paired with oxidation reactions, are processes where an atom or ion gains electrons, leading to a decrease in its oxidation state. This key concept in redox (reduction-oxidation) chemistry showcases how electrons are transferred between species.

A substance that donates electrons serves as a reducing agent, and a substance that accepts electrons acts as an oxidizing agent. In the example provided, manganese dioxide is reduced by carbon, as carbon donates electrons, to produce elemental manganese and carbon monoxide, balanced by the equation \(MnO_2 + C \stackrel{\Delta}{\rightarrow} Mn + CO\). The symbol \(\Delta\) often indicates that heat is applied to the reaction. Balancing reduction reactions involves ensuring that the charge as well as the mass is balanced, which may require the use of oxidation states to determine how electrons are shifted.