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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 H₂S.
  • Oxygen becomes O₂.
  • Sulfur dioxide becomes SO₂.
  • Water vapor becomes H₂O.
By translating all components into their chemical formulas, we get more precise equations:
  • (a) Hydrogen sulfide is H₂S, Oxygen is O₂, Sulfur dioxide is SO₂, and Water vapor is H₂O.
  • (b) Potassium chlorate is KClO₃, Potassium chloride is KCl, and Potassium perchlorate is KClO₄.
  • (c) Hydrogen gas is H₂, Iron(III) oxide is Fe₂O₃, Iron is Fe, and Water vapor is H₂O.
  • (d) Ethane is C₂H₆, Oxygen is O₂, Carbon dioxide is CO₂, and Water vapor is H₂O.
  • (e) Iron(II) chloride is FeCl₂, Chlorine trifluoride is ClF₃, Iron(III) fluoride is FeF₃, and Chlorine gas is Cl₂.
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) H₂S + O₂ → SO₂ + H₂O.
  • (b) KClO₃ → KCl + KClO₄.
  • (c) H₂ + Fe₂O₃ → Fe + H₂O.
  • (d) C₂H₆ + O₂ → CO₂ + H₂O.
  • (e) FeCl₂ + ClF₃ → FeF₃ + Cl₂.
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) H₂S + 3O₂ → SO₂ + 2H₂O.
  • (b) 4KClO₃ → 3KCl + KClO₄.
  • (c) 3H₂ + Fe₂O₃ → 2Fe + 3H₂O.
  • (d) 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
  • (e) 2FeCl₂ + 3ClF₃ → 2FeF₃ + 3Cl₂.
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

Calculate each of the following quantities: (a) Total number of ions in \(38.1 \mathrm{~g}\) of \(\mathrm{SrF}_{2}\) (b) Mass (kg) of \(3.58 \mathrm{~mol}\) of \(\mathrm{CuCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (c) Mass (mg) of \(2.88 \times 10^{22}\) formula units of \(\mathrm{Bi}\left(\mathrm{NO}_{3}\right)_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Fluorine is so reactive that it forms compounds with several of the noble gases. (a) When \(0.327 \mathrm{~g}\) of platinum is heated in fluorine, \(0.519 \mathrm{~g}\) of a dark red, volatile solid forms. What is its empirical formula? (b) When \(0.265 \mathrm{~g}\) of this red solid reacts with excess xenon gas, \(0.378 \mathrm{~g}\) of an orange-yellow solid forms. What is the empirical formula of this compound, the first to contain a noble gas? (c) Fluorides of xenon can be formed by direct reaction of the elements at high pressure and temperature. Under conditions that produce only the tetra- and hexafluorides, \(1.85 \times 10^{-4} \mathrm{~mol}\) of xenon reacted with \(5.00 \times 10^{-4} \mathrm{~mol}\) of fluorine, and \(9.00 \times 10^{-6} \mathrm{~mol}\) of xenon was found in excess. What are the mass percents of each xenon fluoride in the product mixture?

Cocaine \(\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{O}_{4} \mathrm{~N}\right)\) is a natural substance found in \(\operatorname{coca}\) leaves, which have been used for centuries as a local anesthetic and stimulant. Illegal cocaine arrives in the United States either as the pure compound or as the hydrochloride salt \(\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{O}_{4} \mathrm{NHCl}\right)\). At \(25^{\circ} \mathrm{C},\) the salt is very soluble in water \((2.50 \mathrm{~kg} / \mathrm{L}),\) but cocaine is much less so \((1.70 \mathrm{~g} / \mathrm{L})\) (a) What is the maximum mass (g) of the hydrochloride salt that can dissolve in \(50.0 \mathrm{~mL}\) of water? (b) If the solution from part (a) is treated with \(\mathrm{NaOH}\), the salt is converted to cocaine. How much more water (L) is needed to dissolve it?

Percent yields are generally calculated from masses. Would the result be the same if amounts (mol) were used instead? Why?

Aspirin (acetylsalicylic acid, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) ) is made by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left[\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\right]:\) $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}(s)+\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+\mathrm{CH}_{3} \mathrm{COOH}(l) $$ In one preparation, \(3.077 \mathrm{~g}\) of salicylic acid and \(5.50 \mathrm{~mL}\) of acetic anhydride react to form \(3.281 \mathrm{~g}\) of aspirin. (a) Which is the limiting reactant (the density of acetic anhydride is \(1.080 \mathrm{~g} / \mathrm{mL}) ?\) (b) What is the percent yield of this reaction? (c) What is the percent atom economy of this reaction?

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