Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 12

Balance the following equations: (a) $\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{HPO}_{3}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(l)$ (b) $\mathrm{Au}_{2} \mathrm{~S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Au}(s)+\mathrm{H}_{2} \mathrm{~S}(g)$ (c) $\mathrm{Ba}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(a q) \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{NH}_{3}(g)$ (d) $\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$

Short Answer

Expert verified
The balanced equations are: (a) $\mathrm{HClO}_{4}(aq) + \mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(aq) + 2\mathrm{Cl}_{2}\mathrm{O}_{7}(l)$ (b) $\mathrm{Au}_{2}\mathrm{S}_{3}(s) + 3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s) + 3\mathrm{H}_{2}\mathrm{S}(g)$ (c) $\mathrm{Ba}_{3}\mathrm{N}_{2}(s) + 6\mathrm{H}_{2}\mathrm{O}(aq) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(aq) + 2\mathrm{NH}_{3}(g)$ (d) $\mathrm{Na}_{2}\mathrm{CO}_{3}(aq) + 2\mathrm{HCl}(aq) \longrightarrow 2\mathrm{NaCl}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{CO}_{2}(g)$

Step by step solution

01

Since there is 1 Hydrogen atom on the left side, we need to place a coefficient of 4 before HPO3 on the right side of the equation: $\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(a q)+\mathrm{Cl}_{2}\mathrm{O}_{7}(l)$ #\tag_title# Step 2: Balance the number of Chlorine (Cl) atoms

Now we have 4 chlorine atoms on the left side and 2 on the right side. To balance the number of Cl atoms, we need to place a coefficient of 2 before \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) on the right side of the equation: $\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4}\mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(a q)+2\mathrm{Cl}_{2}\mathrm{O}_{7}(l)$ #\tag_title# Step 3: Check the O atoms
02

At this point, we have 16 Oxygen atoms on both sides of the equation, so this equation is now balanced. (b) Balancing the equation: #\tag_title# Step 1: Balance the number of Au atoms

Since there are 2 Au atoms on the left side, we need to place a coefficient of 2 before \(\mathrm{Au}\) on the right side of the equation: $\mathrm{Au}_{2}\mathrm{S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s)+\mathrm{H}_{2}\mathrm{S}(g)$ #\tag_title# Step 2: Balance the number of S atoms
03

Now we have 3 S atoms on the left side and 1 on the right side. To balance the number of S atoms, we need to place a coefficient of 3 before \(\mathrm{H}_{2}\mathrm{S}\) on the right side of the equation: $\mathrm{Au}_{2}\mathrm{S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s)+3\mathrm{H}_{2}\mathrm{S}(g)$ #\tag_title# Step 3: Balance the number of H atoms

Now we have 2 H atoms on the left side and 6 on the right side. To balance the number of H atoms, we need to place a coefficient of 3 before \(\mathrm{H}_{2}\) on the left side of the equation: $ \mathrm{Au}_{2}\mathrm{S}_{3}(s)+3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s)+3\mathrm{H}_{2}\mathrm{S}(g)$ (c) Balancing the equation: #\tag_title# Step 1: Balance the number of Ba atoms
04

Since there are 3 Ba atoms on the left side, we need to place a coefficient of 3 before \(\mathrm{Ba}(\mathrm{OH})_{2}\) on the right side of the equation: $\mathrm{Ba}_{3}\mathrm{N}_{2}(s)+\mathrm{H}_{2}\mathrm{O}(a q) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{NH}_{3}(g)$ #\tag_title# Step 2: Balance the number of N atoms

Now we have 2 N atoms on the left side and 1 on the right side. To balance the number of N atoms, we need to place a coefficient of 2 before \(\mathrm{NH}_{3}\) on the right side of the equation: $\mathrm{Ba}_{3}\mathrm{N}_{2}(s)+\mathrm{H}_{2}\mathrm{O}(a q) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(a q)+2\mathrm{NH}_{3}(g)$ #\tag_title# Step 3: Balance the number of H atoms
05

Now we have 2 H atoms on the left side and 12 on the right side. To balance the number of H atoms, we need to place a coefficient of 6 before \(\mathrm{H}_{2}\mathrm{O}\) on the left side of the equation: $\mathrm{Ba}_{3}\mathrm{N}_{2}(s)+6\mathrm{H}_{2}\mathrm{O}(a q) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(a q)+2\mathrm{NH}_{3}(g)$ (d) Balancing the equation: #\tag_title# Step 1: Balance the number of Na atoms

Since there are 2 Na atoms on the left side, we need to place a coefficient of 2 before \(\mathrm{NaCl}\) on the right side of the equation: $\mathrm{Na}_{2}\mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow 2\mathrm{NaCl}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)$ #\tag_title# Step 2: Balance the number of Cl atoms
06

Now we have 1 Cl atom on the left side and 2 on the right side. To balance the number of Cl atoms, we need to place a coefficient of 2 before \(\mathrm{HCl}\) on the left side of the equation: $\mathrm{Na}_{2}\mathrm{CO}_{3}(a q)+2\mathrm{HCl}(a q) \longrightarrow 2\mathrm{NaCl}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)$ #\tag_title# Step 3: Balance the number of H atoms

Now we have 4 H atoms on the left side and 2 on the right side. To balance the number of H atoms, we need to place a coefficient of 2 before \(\mathrm{H}_{2}\mathrm{O}\) on the right side of the equation: $\mathrm{Na}_{2}\mathrm{CO}_{3}(a q)+2\mathrm{HCl}(a q) \longrightarrow 2\mathrm{NaCl}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{CO}_{2}(g)$. All the equations are now balanced.

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

Determine the formula weights of each of the following compounds: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) which is responsible for the rotten smell of spoiled food; (b) sodium perborate, \(\mathrm{NaBO}_{3}\), a substance used as bleach; (c) calcium carbonate, \(\mathrm{CaCO}_{3},\) a substance found in marble. (c) $\mathrm{CF}_{2} \mathrm{Cl}_{2},\( a refrigerant known as Freon; \)(\mathbf{d}) \mathrm{NaHCO}_{3},$ known as baking soda and used in bread and pastry baking; \((\mathbf{e})\) iron pyrite, \(\mathrm{FeS}_{2}\) which has a golden appearance and is known as "Fool's Gold."

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{NaClO}_{3}(s) \longrightarrow \mathrm{NaCl}(s)+\mathrm{O}_{2}(g)\) (b) $\mathrm{NH}_{4} \mathrm{OH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g)$ (c) \(\mathrm{K}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s)\) (d) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(l)$

Write a balanced chemical equation for the reaction that occurs when (a) \(\mathrm{Mg}(s)\) reacts with \(\mathrm{Cl}_{2}(g) ;(\mathbf{b})\) barium carbonate decomposes into barium oxide and carbon dioxide gas when heated; \((\mathbf{c})\) the hydrocarbon styrene, \(\mathrm{C}_{8} \mathrm{H}_{8}(l),\) is combusted in air; \((\mathbf{d})\) dimethylether, $\mathrm{CH}_{3} \mathrm{OCH}_{3}(g),$ is combusted in air.

The molecular formula of saccharin, an artificial sweetener, is \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S} .(\mathbf{a})\) What is the molar mass of saccharin? (b) How many moles of sachharin are in $2.00 \mathrm{mg}\( of this substance? (c) How many molecules are in \)2.00 \mathrm{mg}\( of this substance? (d) How many C atoms are present in \)2.00 \mathrm{mg}$ of saccharin?

(a) What is the mass, in grams, of one mole of \({ }^{79} \mathrm{Br}\) ? (b) How many bromine atoms are present in one mole of \({ }^{79} \mathrm{Br}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

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