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Chapter 3: Problem 109

Convert the following descriptions into balanced equations: (a) In a gaseous reaction, hydrogen sulfide burns in oxygen to form sulfur dioxide and water vapor. (b) When crystalline potassium chlorate is heated to just above its melting point, it reacts to form two different crystalline compounds, potassium chloride and potassium perchlorate. (c) When hydrogen gas is passed over powdered iron(III) oxide, iron metal and water vapor form. (d) The combustion of gaseous ethane in air forms carbon dioxide and water vapor. (e) Iron(II) chloride is converted to iron(III) fluoride by treatment with chlorine trifluoride gas. Chlorine gas is also formed.

Short Answer

Expert verified
(a) H₂S + 3O₂ → SO₂ + 2H₂O, (b) 4KClO₃ → 3KCl + KClO₄, (c) 3H₂ + Fe₂O₃ → 2Fe + 3H₂O, (d) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, (e) 2FeCl₂ + 3ClF₃ → 2FeF₃ + 3Cl₂

Step by step solution

01

Write Word Equations

Convert each description into a word equation.(a) Hydrogen sulfide + Oxygen → Sulfur dioxide + Water vapor(b) Potassium chlorate → Potassium chloride + Potassium perchlorate(c) Hydrogen gas + Iron(III) oxide → Iron + Water vapor(d) Ethane + Oxygen → Carbon dioxide + Water vapor(e) Iron(II) chloride + Chlorine trifluoride → Iron(III) fluoride + Chlorine gas
02

Write Chemical Formulas

Replace the names with their chemical formulas.(a) Hydrogen sulfide is H₂S, Oxygen is O₂, Sulfur dioxide is SO₂, Water vapor is H₂O.(b) Potassium chlorate is KClO₃, Potassium chloride is KCl, Potassium perchlorate is KClO₄.(c) Hydrogen gas is H₂, Iron(III) oxide is Fe₂O₃, Iron is Fe, Water vapor is H₂O.(d) Ethane is C₂H₆, Oxygen is O₂, Carbon dioxide is CO₂, Water vapor is H₂O.(e) Iron(II) chloride is FeCl₂, Chlorine trifluoride is ClF₃, Iron(III) fluoride is FeF₃, Chlorine gas is Cl₂.
03

Write Unbalanced Equations

Convert the word equations with chemical formulas into unbalanced equations.(a) H₂S + O₂ → SO₂ + H₂O(b) KClO₃ → KCl + KClO₄(c) H₂ + Fe₂O₃ → Fe + H₂O(d) C₂H₆ + O₂ → CO₂ + H₂O(e) FeCl₂ + ClF₃ → FeF₃ + Cl₂
04

Balance Each Equation

Ensure the number of each type of atom is the same on both sides of the equation.(a) H₂S + 3O₂ → SO₂ + 2H₂O(b) 4KClO₃ → 3KCl + KClO₄(c) 3H₂ + Fe₂O₃ → 2Fe + 3H₂O(d) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O(e) 2FeCl₂ + 3ClF₃ → 2FeF₃ + 3Cl₂

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

word equations
Chemical reactions can be communicated using word equations. These equations are straightforward sentences that describe the reactants and products of a reaction. For example, 'Hydrogen sulfide reacts with oxygen to form sulfur dioxide and water vapor.' The word equation simplifies this to: Hydrogen sulfide + Oxygen → Sulfur dioxide + Water vapor.
Word equations are essential because they help you understand the basics of what's happening without diving into complex formulas. For instance:
  • (a) Hydrogen sulfide + Oxygen → Sulfur dioxide + Water vapor
  • (b) Potassium chlorate → Potassium chloride + Potassium perchlorate
  • (c) Hydrogen gas + Iron(III) oxide → Iron + Water vapor
  • (d) Ethane + Oxygen → Carbon dioxide + Water vapor
  • (e) Iron(II) chloride + Chlorine trifluoride → Iron(III) fluoride + Chlorine gas
These word equations set the foundation for understanding the chemical changes occurring during a reaction.
chemical formulas
Word equations can be converted into chemical formulas that use symbols and subscripts to represent chemical species. This allows us to quantify the substances more accurately. For example:
  • Hydrogen sulfide becomes Hâ‚‚S.
  • Oxygen becomes Oâ‚‚.
  • Sulfur dioxide becomes SOâ‚‚.
  • Water vapor becomes Hâ‚‚O.
By translating all components into their chemical formulas, we get more precise equations:
  • (a) Hydrogen sulfide is Hâ‚‚S, Oxygen is Oâ‚‚, Sulfur dioxide is SOâ‚‚, and Water vapor is Hâ‚‚O.
  • (b) Potassium chlorate is KClO₃, Potassium chloride is KCl, and Potassium perchlorate is KClOâ‚„.
  • (c) Hydrogen gas is Hâ‚‚, Iron(III) oxide is Feâ‚‚O₃, Iron is Fe, and Water vapor is Hâ‚‚O.
  • (d) Ethane is Câ‚‚H₆, Oxygen is Oâ‚‚, Carbon dioxide is COâ‚‚, and Water vapor is Hâ‚‚O.
  • (e) Iron(II) chloride is FeClâ‚‚, Chlorine trifluoride is ClF₃, Iron(III) fluoride is FeF₃, and Chlorine gas is Clâ‚‚.
This step provides clarity on the elemental composition of reactants and products, making the equations useful for further analysis and balancing.
unbalanced equations
After converting word equations to chemical formulas, the next step is creating unbalanced equations. Unbalanced equations show the raw chemical elements involved but do not account for the law of conservation of mass. This law states that matter is neither created nor destroyed in a chemical reaction.
Unbalanced equations look like:
  • (a) Hâ‚‚S + Oâ‚‚ → SOâ‚‚ + Hâ‚‚O.
  • (b) KClO₃ → KCl + KClOâ‚„.
  • (c) Hâ‚‚ + Feâ‚‚O₃ → Fe + Hâ‚‚O.
  • (d) Câ‚‚H₆ + Oâ‚‚ → COâ‚‚ + Hâ‚‚O.
  • (e) FeClâ‚‚ + ClF₃ → FeF₃ + Clâ‚‚.
These equations highlight the components of the reaction but lack balance in both the number and types of atoms on each side. Recognizing an unbalanced equation is key. It sets the stage for correct balancing later on.
balanced equations
Balancing chemical equations is vital in reflecting the conservation of mass. A balanced equation ensures that the number of atoms of each element is the same on both sides of the equation.
To balance, adjust coefficients (the numbers in front of compounds) without altering the chemical formulas:
  • (a) Hâ‚‚S + 3Oâ‚‚ → SOâ‚‚ + 2Hâ‚‚O.
  • (b) 4KClO₃ → 3KCl + KClOâ‚„.
  • (c) 3Hâ‚‚ + Feâ‚‚O₃ → 2Fe + 3Hâ‚‚O.
  • (d) 2Câ‚‚H₆ + 7Oâ‚‚ → 4COâ‚‚ + 6Hâ‚‚O.
  • (e) 2FeClâ‚‚ + 3ClF₃ → 2FeF₃ + 3Clâ‚‚.
Balancing maintains atomic parity and validates the equation based on physical realities. This makes the chemical equation meaningful and useful for predicting the outcome and quantities in a reaction. Understanding the steps involved in balancing equations will strengthen your grasp of chemical processes and reactions.

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Most popular questions from this chapter

Ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuric acid, ethanol forms diethyl ether and water: $$ 2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(g) $$ In a side reaction, some ethanol forms ethylene and water: $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{2} \mathrm{CH}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) If \(50.0 \mathrm{~g}\) of ethanol yields \(35.9 \mathrm{~g}\) of diethyl ether, what is the percent yield of diethyl ether? (b) If \(45.0 \%\) of the ethanol that did not produce the ether reacts by the side reaction, what mass (g) of ethylene is produced?

Potassium nitrate decomposes on heating, producing potassium oxide and gaseous nitrogen and oxygen: $$ 4 \mathrm{KNO}_{3}(s) \longrightarrow 2 \mathrm{~K}_{2} \mathrm{O}(s)+2 \mathrm{~N}_{2}(g)+5 \mathrm{O}_{2}(g) $$ To produce \(56.6 \mathrm{~kg}\) of oxygen, how many (a) moles of \(\mathrm{KNO}_{3}\) and (b) grams of \(\mathrm{KNO}_{3}\) must be heated?

Two successive reactions, \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{C},\) have yields of \(73 \%\) and \(68 \%,\) respectively. What is the overall percent yield for conversion of A to C?

The multistep smelting of ferric oxide to form elemental iron occurs at high temperatures in a blast furnace. In the first step, ferric oxide reacts with carbon monoxide to form \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). This substance reacts with more carbon monoxide to form iron(II) oxide, which reacts with still more carbon monoxide to form molten iron. Carbon dioxide is also produced in each step. (a) Write an overall balanced equation for the iron- smelting process. (b) How many grams of carbon monoxide are required to form 45.0 metric tons of iron from ferric oxide?

Hemoglobin is \(6.0 \%\) heme \(\left(\mathrm{C}_{34} \mathrm{H}_{32} \mathrm{FeN}_{4} \mathrm{O}_{4}\right)\) by mass. To remove the heme, hemoglobin is treated with acetic acid and \(\mathrm{NaCl},\) which forms hemin \(\left(\mathrm{C}_{34} \mathrm{H}_{32} \mathrm{~N}_{4} \mathrm{O}_{4} \mathrm{FeCl}\right) .\) A blood sample from a crime scene contains \(0.65 \mathrm{~g}\) of hemoglobin. (a) How many grams of heme are in the sample? (b) How many moles of heme? (c) How many grams of Fe? (d) How many grams of hemin could be formed for a forensic chemist to measure?
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