Log out Go to App
Go to App

Learning Materials

Features

Discover

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A piece of aluminum foil \(1.00 \mathrm{~cm}\) square and \(0.550 \mathrm{~mm}\) thick is allowed to react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (The density of aluminum is \(\left.2.699 \mathrm{~g} / \mathrm{cm}^{3} .\right)\) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completelv?

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) Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield? (c) Can a reaction ever have \(110 \%\) actual yield?

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 \(=88 \mathrm{~g} / \mathrm{mol}\).

Sodium hydroxide reacts with carbon dioxide as follows: \(2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and 1.00 \(\mathrm{mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free