The following chemical equation is unbalanced: \(\mathrm{I}_{2}+\mathrm{Cl}_{2} \rightarrow \mathrm{ICl}_{3}\) (a) Balance the equation. (b) The balanced equation is a source of many conversion factors containing the two reactants and one product. Write all of them that use the word mole(s). (c) The balanced equation is also a source of conversion factors involving the individual atoms and molecules. Write all of them that use the word mole(s).

Short Answer

Expert verified
(a) The balanced chemical equation is \(\mathrm{I}_{2} + 3\mathrm{Cl}_{2} \rightarrow 2\mathrm{ICl}_{3}\). (b) Mole-based conversion factors: 1. \(\frac{1 \text{ mole of } \mathrm{I}_2}{3 \text{ moles of } \mathrm{Cl}_2}\) 2. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{1 \text{ mole of } \mathrm{I}_2}\) 3. \(\frac{1 \text{ mole of } \mathrm{I}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 4. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{1 \text{ mole of } \mathrm{I}_2}\) 5. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 6. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{3 \text{ moles of } \mathrm{Cl}_2}\) (c) Atom-based conversion factors: 1. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of Cl atoms}}\) 2. \(\frac{6 \text{ moles of Cl atoms}}{2 \text{ moles of I atoms}}\) 3. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of ICl molecules}}\) 4. \(\frac{6 \text{ moles of ICl molecules}}{2 \text{ moles of I atoms}}\) 5. \(\frac{6 \text{ moles of Cl atoms}}{4 \text{ moles of ICl molecules}}\) 6. \(\frac{4 \text{ moles of ICl molecules}}{6 \text{ moles of Cl atoms}}\)

Step by step solution

01

Write the given (unbalanced) chemical equation

The given chemical equation is: \[\mathrm{I}_{2}+\mathrm{Cl}_{2} \rightarrow \mathrm{ICl}_{3}\]
02

Identify the elements in the equation

We have the following elements in the equation: - Iodine (I) - Chlorine (Cl)
03

Balance the equation for each element

Balance each element by adjusting the coefficients in front of molecules/compounds. Iodine (I): - There are 2 iodine atoms on the left side (LHS) and 2 iodine atoms in 1 mole of \(\mathrm{ICl}_3\) on the right side (RHS). So Iodine is already balanced. Chlorine (Cl): - There are 2 chlorine atoms in 1 mole of \(\mathrm{Cl}_2\) on the left side and 3 chlorine atoms in 1 mole of \(\mathrm{ICl}_3\) on the right side. To balance the Cl atoms, we need to find the least common multiple (LCM) of 2 and 3 which is 6. We can achieve this by putting coefficients of 3 in front of \(\mathrm{Cl}_2\) and 2 in front of \(\mathrm{ICl}_3\): \[ \mathrm{I}_2 + 3\mathrm{Cl}_2 \rightarrow 2\mathrm{ICl}_3\] Now we have a balanced chemical equation. #b. Write all conversion factors containing the two reactants and one product.#
04

Find all mole-based conversion factors

Using the balanced chemical equation, we can find mole-based conversion factors: 1. \(\frac{1 \text{ mole of } \mathrm{I}_2}{3 \text{ moles of } \mathrm{Cl}_2}\) 2. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{1 \text{ mole of } \mathrm{I}_2}\) 3. \(\frac{1 \text{ mole of } \mathrm{I}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 4. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{1 \text{ mole of } \mathrm{I}_2}\) 5. \(\frac{3 \text{ moles of } \mathrm{Cl}_2}{2 \text{ moles of } \mathrm{ICl}_3}\) 6. \(\frac{2 \text{ moles of } \mathrm{ICl}_3}{3 \text{ moles of } \mathrm{Cl}_2}\) #c. Write all conversion factors involving individual atoms and molecules.#
05

Find atom-based conversion factors

We can find atom-based conversion factors using individual atoms from the balanced equation. 1. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of Cl atoms}}\) 2. \(\frac{6 \text{ moles of Cl atoms}}{2 \text{ moles of I atoms}}\) 3. \(\frac{2 \text{ moles of I atoms}}{6 \text{ moles of ICl molecules}}\) 4. \(\frac{6 \text{ moles of ICl molecules}}{2 \text{ moles of I atoms}}\) 5. \(\frac{6 \text{ moles of Cl atoms}}{4 \text{ moles of ICl molecules}}\) 6. \(\frac{4 \text{ moles of ICl molecules}}{6 \text{ moles of Cl atoms}}\)

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.

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

If you have \(0.262\) mole of iron(III) sulfate, how many oxygen atoms do you have?

Vitamin \(\mathrm{C}\), also known as ascorbic acid, contains carbon, hydrogen, and possibly oxygen. A \(0.160 \mathrm{~g}\) sample of ascorbic acid is subjected to combustion analysis, yielding \(40.93 \%\) C and \(4.58 \%\) H. If the molar mass of ascorbic acid is approximately \(176 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with fluorine gas \(\left(\mathrm{F}_{2}\right)\) to form carbon tetrafluoride gas and hydrogen fluoride gas. If \(2.78 \mathrm{~g}\) of ethylene reacted with an excess of fluorine, how many grams of each product could theoretically be produced?

Butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), used as the fuel in disposable lighters, reacts with oxygen \(\left(\mathrm{O}_{2}\right)\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Suppose \(10.00 \mathrm{~g}\) butane is combined with \(10.00 \mathrm{~g} \mathrm{O}_{2}\). (a) Write a balanced chemical equation for the combustion of butane. (b) Which reactant is limiting? (c) What is the theoretical yield of each product for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) How many additional grams of the limiting reactant are required to run this reaction in a balanced fashion?

Copper(I) oxide reacts with solid carbon to form copper metal. Carbon dioxide gas is the other product of this reaction. (a) Write the balanced chemical equation for this reaction. (b) Coke is a cheap, impure form of solid carbon that is often used industrially. If a sample of coke is \(95 \%\) C by mass, determine the mass in kilograms of coke needed to react completely with \(1.000\) ton of copper(I) oxide. \([1\) ton \(=2000 \mathrm{lb} ; 1 \mathrm{~kg}=2.205 \mathrm{lb}]\)

See all solutions

Recommended explanations on Chemistry Textbooks

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