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Write a balanced equation for the reaction that occurs in each of the following cases: (a) Ozone decomposes to dioxygen. (b) Xenon reacts with fluorine. (Write three different equations.) (c) Sulfur reacts with hydrogen gas. (d) Fluorine reacts with water.

Short Answer

Expert verified
a) 2O3 → 3O2 b) 1) Xe + 3F2 → XeF6 2) Xe + 2F2 → XeF4 3) Xe + F2 → XeF2 c) S + H2 → H2S d) F2 + 2H2O → 4HF + O2

Step by step solution

01

Write unbalanced equation

Ozone (O3) decomposes to dioxygen (O2), which can be written as: O3 → O2
02

Balance the equation

In order to balance the equation, we need two molecules of ozone decomposing to three molecules of dioxygen, so the balanced equation is: 2O3 → 3O2 b) Xenon reacts with fluorine (Three different equations).
03

Write unbalanced equations

Xenon can react with fluorine to form three different xenon fluorides: Xenon hexafluoride (XeF6), xenon tetrafluoride (XeF4), and xenon difluoride (XeF2). The three unbalanced reactions are: 1) Xe + F2 → XeF6 2) Xe + F2 → XeF4 3) Xe + F2 → XeF2
04

Balance the equations

To balance these three reactions, we must adjust the stoichiometric coefficients like this: 1) Xe + 3F2 → XeF6 2) Xe + 2F2 → XeF4 3) Xe + F2 → XeF2 c) Sulfur reacts with hydrogen gas.
05

Write unbalanced equation

Sulfur (S) reacts with hydrogen gas (H2) to form hydrogen sulfide (H2S), which can be written as: S + H2 → H2S
06

Balance the equation

In order to balance the equation, we need two hydrogen atoms for every one sulfur atom. The balanced equation is: S + H2 → H2S d) Fluorine reacts with water.
07

Write unbalanced equation

Fluorine (F2) reacts with water (H2O) to form hydrogen fluoride (HF) and oxygen gas (O2), which can be written as: F2 + H2O → HF + O2
08

Balance the equation

In order to balance the equation, we must have four atoms of hydrogen, two atoms of oxygen and two atoms of fluorine on both sides. So, the balanced equation is: F2 + 2H2O → 4HF + O2

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Key Concepts

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

Balancing Equations
Balancing chemical equations is all about ensuring that the number of atoms for each element is the same on both sides of the reaction. This makes sure that mass is conserved during the reaction, a fundamental principle in chemistry.
To balance an equation, follow these steps:
  • Write the unbalanced equation.
  • Count the number of each type of atom on both sides of the equation.
  • Add coefficients to the compounds to equalize the number of atoms per element on each side.
  • Double-check your work to ensure mass is conserved.
For instance, when ozone decomposes to dioxygen, the unbalanced equation is written as:
\( \text{O}_3 \rightarrow \text{O}_2 \).
In this case, you will balance it by adding a coefficient of 2 before \( \text{O}_3 \) and a 3 before \( \text{O}_2 \) resulting in:
\( 2\text{O}_3 \rightarrow 3\text{O}_2 \). This shows that 2 ozone molecules decompose into 3 dioxygen molecules.
Stoichiometry
Stoichiometry is the exciting process of using balanced equations to determine the quantitative relationships between reactants and products in a chemical reaction. It is crucial because it allows chemists to predict how much of a substance is needed or produced in a given reaction.
For example, in the balanced reaction between xenon and fluorine to form XeF6: \( \text{Xe} + 3\text{F}_2 \rightarrow \text{XeF}_6 \), you can see that 1 mole of xenon reacts with 3 moles of fluorine to produce 1 mole of xenon hexafluoride. From here, stoichiometry can help calculate the amounts of each reactant required to obtain a specified amount of product, and vice versa.
A helpful tip is to always use the mole ratio from the balanced equation to convert between different chemical species.
Reaction Types
Understanding different types of chemical reactions can help in predicting the outcomes and behavior of reactions. Here are some common types:
  • Decomposition Reaction: A single compound breaks down into simpler substances, like the decomposition of ozone to dioxygen.
  • Synthesis Reaction: Simple substances combine to form a more complex compound, like sulfur reacting with hydrogen to produce hydrogen sulfide.
  • Single Replacement Reaction: An element replaces a similar element in a compound.
  • Double Replacement Reaction: Two compounds exchange partners to form two new compounds.
For instance, the reaction of fluorine with water is more of a synthesis type, where new products (HF and \( \text{O}_2 \)) are formed from reactants.
Moreover, recognizing the reaction type helps in balancing the equation and predicting the direction and products of the reaction.

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Most popular questions from this chapter

Write balanced equations for the following reactions: (a) barium oxide with water, (b) iron(II) oxide with perchloric acid, (c) sulfur trioxide with water, (d) carbon dioxide with aqueous sodium hydroxide.

Some ions do not have a corresponding neutral atom that has the same electron configuration. For each of the following ions, identify the neutral atom that has the same number of electrons and determine if this atom has the same electron configuration. If such an atom does not exist, explain why. (b) \(\mathrm{Sc}^{3+}\) (d) \(\mathrm{Zn}^{2+},(\mathrm{e}) \mathrm{Sn}^{4+}\) (a) \(\mathrm{Cl}\) (c) \(\mathrm{Fe}^{2+}\)

Explain the following variations in atomic or ionic radii: (a) \(\Gamma>\mathrm{I}>\mathrm{I}^{+},(\mathrm{b}) \mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

For each of the following pairs, indicate which element has the smaller first ionization energy: (a) \(\mathrm{Ti}, \mathrm{Ba} ;(\mathrm{b}) \mathrm{Ag}, \mathrm{Cu} ;(\mathrm{c}) \mathrm{Ge}\) \(\mathrm{Cl} ;\) (d) \(\mathrm{Pb},\) Sb. (In each case use electron configuration and effective nuclear charge to explain your answer.)

Mercury in the environment can exist in oxidation states 0,+1 , and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.107 ), but instead of using ultraviolet light to eject valence electrons, X-rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV},\) from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the 1 s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for \(\mathrm{O} .\) (a) Calculate the wavelength of the X-rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the 1 s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full + \(1 .\)

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