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Write balanced equations for the formation of the following compounds from their elements: (a) iron(III) oxide (b) sucrose (table sugar, \(\left.\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (c) uranium hexafluoride (a solid at \(25^{\circ} \mathrm{C}\) )

Short Answer

Expert verified
(a) 4 Fe + 3 O₂ → 2 Fe₂O₃; (b) 12 C + 11 H₂ + 11 O₂ → 2 C₁₂H₂₂O₁₁; (c) U + 3 F₂ → UF₆.

Step by step solution

01

Identify the Elements

To form each compound, we need to identify the element reactants for each compound: (a) Iron(III) oxide is composed of iron (Fe) and oxygen (O). (b) Sucrose consists of carbon (C), hydrogen (H), and oxygen (O). (c) Uranium hexafluoride is composed of uranium (U) and fluorine (F).
02

Write Unbalanced Equations

Write the chemical equations using the identified elements: (a) Fe(s) + O₂(g) → Fe₂O₃(s) (b) C(s) + H₂(g) + O₂(g) → C₁₂H₂₂O₁₁(s) (c) U(s) + F₂(g) → UF₆(s)
03

Balance Iron(III) Oxide Equation

For Fe₂O₃, the equation is balanced as follows: Start with Fe: 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s). Each side has 4 iron and 6 oxygen atoms.
04

Balance Sucrose Equation

For sucrose, balance the equation: 12 C(s) + 11 H₂(g) + rac{11}{2} O₂(g) → C₁₂H₂₂O₁₁(s). Adjust all coefficients to whole numbers: 12 C(s) + 11 H₂(g) + 11 O₂(g) → 2 C₁₂H₂₂O₁₁(s). Each side now has 24 carbon, 22 hydrogen, and 22 oxygen atoms.
05

Balance Uranium Hexafluoride Equation

For UF₆, the equation is balanced as follows: U(s) + 3 F₂(g) → UF₆(s). Each side has 1 uranium and 6 fluorine atoms.

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is an important step in demonstrating the conservation of mass in a chemical reaction. Ensuring that there is an equal number of each type of atom on both the reactant and product sides is paramount. This means rewriting the chemical reaction so that it satisfies this fundamental law. Let's break it down:
  • Start by writing the unbalanced equation using the chemical symbols of the reactants and the product.
  • Count how many atoms of each element exist in the reactants and products.
  • Adjust the coefficients in front of each compound or element until the number of each type of atom is equal on both sides.
For example, with iron(III) oxide, the balanced equation shows 4 Fe and 6 O atoms on each side: \[ 4 \text{Fe(s)} + 3 \text{O}_2(\text{g}) \rightarrow 2 \text{Fe}_2\text{O}_3(\text{s}) \]Remember, only the coefficients are changed when balancing equations, never the subscripts within a chemical formula. This maintains the compound's correct composition.
Elemental Composition
Elemental composition refers to the types and number of atoms that make up a compound. Each compound has a unique combination of elements. Understanding this composition is crucial for writing and balancing chemical equations.
Take iron(III) oxide, for example:
  • It consists of iron (Fe) and oxygen (O) atoms.
  • Its formula is \( \text{Fe}_2\text{O}_3 \), indicating it contains two iron atoms and three oxygen atoms per molecule.
Now look at sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)):
  • It includes carbon (C), hydrogen (H), and oxygen (O) atoms.
  • The formula implies it has 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms per molecule.
Understanding these basic building blocks of compounds ensures accuracy in chemical equations and provides insight into the compound's properties.
Compound Formation
Compound formation involves combining elements in specific ratios to create new chemical substances. This process aligns with the laws of chemical combination. Compounds form with definitive proportions of their constituent elements, decided by their chemical formulas.
For example, in forming uranium hexafluoride, uranium atoms combine with fluorine atoms:
  • The balanced chemical formula \( \text{UF}_6 \) shows that one uranium combines with six fluorine atoms.
  • This precise stoichiometry results in the compound's stable structure and properties.
Such precise measurements and ratios are fundamental in industrial applications, research, and everyday chemistry. Understanding how elements come together to form compounds gives us insights into chemical reactivity and synthesis.

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Most popular questions from this chapter

The reaction \(\mathrm{S}_{8}(\mathrm{~g}) \longrightarrow 4 \mathrm{~S}_{2}(\mathrm{~g})\) has \(\Delta H^{\circ}=+237 \mathrm{k}\) ) (a) The \(\mathrm{S}_{8}\) molecule has eight sulfur atoms arranged in a ring. What is the hybridization and geometry around each sulfur atom in \(\mathrm{S}_{8}\) ? (b) The average \(\mathrm{S}-\mathrm{S}\) bond dissociation energy is \(225 \mathrm{~kJ} / \mathrm{mol}\). Using the value of \(\Delta H^{\circ}\) given above, what is the \(\mathrm{S}=\mathrm{S}\) double bond energy in \(\mathrm{S}_{2}(g)\) ? (c) Assuming that the bonding in \(\mathrm{S}_{2}\) is similar to the bonding in \(\mathrm{O}_{2}\), give a molecular orbital description of the bonding in \(\mathrm{S}_{2}\). Is \(S_{2}\) likely to be paramagnetic or diamagnetic?

Assume that a particular reaction evolves \(244 \mathrm{~kJ}\) of heat and that \(35 \mathrm{~kJ}\) of \(P V\) work is gained by the system. What are the values of \(\Delta E\) and \(\Delta H\) for the system? For the surroundings?

Calculate the amount of heat required to raise the temperature of \(250.0 \mathrm{~g}\) (approximately 1 cup) of hot chocolate from \(25.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C}\). Assume hot chocolate has the same specific heat as water \(\left[4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\)

One of the steps in the cracking of petroleum into gasoline in. volves the thermal breakdown of large hydrocarbon molecules into smaller ones. For example, the following reaction might occur: $$\mathrm{C}_{11} \mathrm{H}_{24} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{C}_{4} \mathrm{H}_{8}+\mathrm{C}_{3} \mathrm{H}_{6}$$ Is \(\Delta S\) for this reaction likely to be positive or negative? Explain.

Assume that \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{CsOH}\) and \(50.0 \mathrm{~mL}\) of \(0.400\) M HCl are mixed in a calorimeter. The solutions start out at \(22.50^{\circ} \mathrm{C}\), and the final temperature after reaction is \(24.28^{\circ} \mathrm{C}\). The densities of the solutions are all \(1.00 \mathrm{~g} / \mathrm{mL}\), and the specific heat of the mixture is \(4.2 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\). What is the enthalpy change for the neutralization reaction of \(1.00 \mathrm{~mol}\) of \(\mathrm{CsOH}\) in \(\mathrm{kJ}\) ?

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