Question

Balance the following equations: (a) $\mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)$ (b) $\mathrm{TiCl}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{TiO}_{2}(s)+\mathrm{HCl}(a q)$ (c) $\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (d) $\mathrm{AlCl}_{3}(s)+\mathrm{Ca}_{3} \mathrm{~N}_{2}(s) \longrightarrow \mathrm{AlN}(s)+\mathrm{CaCl}_{2}(s)$

Short answer

Balanced equations: (a) 6Li(s) + N₂(g) → 2Li₃N(s) (b) TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 4HCl(aq) (c) 2NH₄NO₃(s) → 2N₂(g) + O₂(g) + 4H₂O(g) (d) 3AlCl₃(s) + Ca₃N₂(s) → 2AlN(s) + 3CaCl₂(s)

Step-by-step solution

(a) First Equation: Balancing Li and N atoms

In the given equation, Li(s) + N₂(g) → Li₃N(s), we need to balance the Li (Lithium) and N (Nitrogen) atoms. We have 2 Nitrogen atoms on the reactant side and 1 Nitrogen atom on the product side. We also have 1 Lithium atom on the reactant side and 3 Lithium atoms on the product side.

(a) Finding the coefficients

To balance the Nitrogen atoms, we can add a coefficient of 1/2 to the product side: Li(s) + N₂(g) → \( \frac{1}{2} \) Li₃N(s). To balance the Lithium atoms, we can add a coefficient of 3 to the reactant side: 3Li(s) + N₂(g) → \( \frac{1}{2} \) Li₃N(s).

(a) Getting rid of the fraction

To make all coefficients whole numbers, we multiply the entire equation by 2: 3Li(s) + N₂(g) → Li₃N(s). The balanced equation is: 6Li(s) + N₂(g) → 2Li₃N(s).

(b) Second Equation: Balancing Ti, Cl, H, and O atoms

In the given equation, TiCl₄(l) + H₂O(l) → TiO₂(s) + HCl(aq), we need to balance the Ti (Titanium), Cl (Chlorine), H (Hydrogen), and O (Oxygen) atoms. We have 4 Cl atoms and 1 Ti atom on the reactant side, and 2 Cl atoms, 1 H atom, and 3 O atoms on the product side.

(b) Finding the coefficients

We start by balancing the Chlorine atoms: add a coefficient of 2 to the HCl on the product side: TiCl₄(l) + H₂O(l) → TiO₂(s) + 2HCl(aq). Now, there are 2 H atoms on both sides, and the Oxygen atoms are already balanced, so the balanced equation is: TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 4HCl(aq).

(c) Third Equation: Balancing N, O, and H atoms

In the given equation, NH₄NO₃(s) → N₂(g) + O₂(g) + H₂O(g), we need to balance the N (Nitrogen), O (Oxygen), and H (Hydrogen) atoms. We have 2 Nitrogen atoms, 4 Oxygen atoms, and 4 Hydrogen atoms on the reactant side, and 2 Nitrogen atoms, 2 Oxygen atoms, and 2 Hydrogen atoms on the product side.

(c) Finding the coefficients

To balance the Oxygen atoms and leave the already balanced Nitrogen atoms unchanged, we add a coefficient of 1/2 to the O₂ on the product side: NH₄NO₃(s) → N₂(g) + \( \frac{1}{2} \) O₂(g) + H₂O(g). Now, there are 4 Hydrogen atoms on the reactant side and only 2 on the product side, so we add a coefficient of 2 to the H₂O on the product side: NH₄NO₃(s) → N₂(g) + \( \frac{1}{2} \) O₂(g) + 2H₂O(g).

(c) Getting rid of the fraction

To make all coefficients whole numbers, we multiply the entire equation by 2: 2NH₄NO₃(s) → 2N₂(g) + O₂(g) + 4H₂O(g). The balanced equation is: 2NH₄NO₃(s) → 2N₂(g) + O₂(g) + 4H₂O(g).

(d) Fourth Equation: Balancing Al, N, Cl, and Ca atoms

In the given equation, AlCl₃(s) + Ca₃N₂(s) → AlN(s) + CaCl₂(s), we need to balance the Al (Aluminum), N (Nitrogen), Cl (Chlorine), and Ca (Calcium) atoms. We have 3 Cl atoms, 1 Al atom, 3 Ca atoms, and 2 N atoms on the reactant side, and 1 Cl atom, 1 Al atom, 1 Ca atom, and 1 N atom on the product side.

(d) Finding the coefficients

To balance the Nitrogen atoms, we add a coefficient of 2 to AlN on the product side: AlCl₃(s) + Ca₃N₂(s) → 2AlN(s) + CaCl₂(s). Now match 3 Calcium atomic ratio by adding a coefficient of 3 to the CaCl₂ on the product side: AlCl₃(s) + Ca₃N₂(s) → 2AlN(s) + 3CaCl₂(s). Finally, balance the Chlorine atoms by adding a coefficient of 3 to the AlCl₃ on the reactant side: 3AlCl₃(s) + Ca₃N₂(s) → 2AlN(s) + 3CaCl₂(s). The balanced equation is: 3AlCl₃(s) + Ca₃N₂(s) → 2AlN(s) + 3CaCl₂(s).