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Chapter 3: Problem 12

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

Expert verified
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

01

(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.
02

(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).
03

(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).
04

(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.
05

(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).
06

(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.
07

(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).
08

(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).
09

(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.
10

(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).

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Most popular questions from this chapter

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2},\) and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

An element \(X\) forms an iodide \(\left(\mathrm{Xl}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{Xl}_{3}\) is treated, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2},\) molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) molar mass \(=88 \mathrm{~g} / \mathrm{mol}\).

Detonation of nitroglycerin proceeds as follows: $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}_{3} \mathrm{O}_{9}(l) \longrightarrow \\ \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad12 \mathrm{CO}_{2}(g)+6 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) If a sample containing 2.00 \(\mathrm{mL}\) of nitroglycerin (density \(=\) 1.592 \(\mathrm{g} / \mathrm{mL}\) ) is detonated, how many moles of gas are produced? (b) If each mole of gas occupies 55 Lunder the conditions of the explosion, how many liters of gas are produced? (c) How many grams of \(\mathrm{N}_{2}\) are produced in the detonation?

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: \(\mathrm{O}_{3}(g)+2 \mathrm{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) $$ (a) How many moles of sodium iodide are needed to remove \(5.95 \times 10^{-6} \mathrm{~mol}\) of \(\mathrm{O}_{3} ?\) (b) How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3} ?\)
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