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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 O鈧 鈫 2 Fe鈧侽鈧; (b) 12 C + 11 H鈧 + 11 O鈧 鈫 2 C鈧佲倐H鈧傗倐O鈧佲倎; (c) U + 3 F鈧 鈫 UF鈧.

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) + O鈧(g) 鈫 Fe鈧侽鈧(s) (b) C(s) + H鈧(g) + O鈧(g) 鈫 C鈧佲倐H鈧傗倐O鈧佲倎(s) (c) U(s) + F鈧(g) 鈫 UF鈧(s)
03

Balance Iron(III) Oxide Equation

For Fe鈧侽鈧, the equation is balanced as follows: Start with Fe: 4 Fe(s) + 3 O鈧(g) 鈫 2 Fe鈧侽鈧(s). Each side has 4 iron and 6 oxygen atoms.
04

Balance Sucrose Equation

For sucrose, balance the equation: 12 C(s) + 11 H鈧(g) + rac{11}{2} O鈧(g) 鈫 C鈧佲倐H鈧傗倐O鈧佲倎(s). Adjust all coefficients to whole numbers: 12 C(s) + 11 H鈧(g) + 11 O鈧(g) 鈫 2 C鈧佲倐H鈧傗倐O鈧佲倎(s). Each side now has 24 carbon, 22 hydrogen, and 22 oxygen atoms.
05

Balance Uranium Hexafluoride Equation

For UF鈧, the equation is balanced as follows: U(s) + 3 F鈧(g) 鈫 UF鈧(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

Is the Haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at \(25^{\circ} \mathrm{C} ?\) At what temperature \(\left({ }^{\circ} \mathrm{C}\right)\) does the changeover occur? \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H^{\circ}=-92.2 \mathrm{~kJ} ; \Delta S^{\circ}=-199 \mathrm{~J} / \mathrm{K}\)

Hydrazine, a component of rocket fuel, undergoes combustion to yield \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Draw an electron-dot structure for hydrazine, predict the geometry about each nitrogen atom, and tell the hybridization of each nitrogen. (b) Use the following information to set up a Hess's law cycle, and then calculate \(\Delta H^{\circ}\) for the combustion reaction. You will need to use fractional coefficients for some equations. $$\begin{aligned}&2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ &\Delta H^{\circ}=-1011.2 \mathrm{~kJ} \\ &\mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \\ &\Delta H^{\circ}=-317.2 \mathrm{~kJ} \\ &4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ &\Delta H^{\circ}=-286.0 \mathrm{~kJ} \\ &\mathrm{H}_{2} \mathrm{O}(I) \quad \Delta H_{\mathrm{f}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ (c) How much heat is released on combustion of \(100.0 \mathrm{~g}\) of hydrazine?

(a) When \(0.350 \mathrm{~g}\) of biodiesel \(\left(\mathrm{C}_{19} \mathrm{H}_{38} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter, the temperature of both the water and the calorimeter rise in temperature by \(6.85^{\circ} \mathrm{C}\). Assuming that the bath contains \(300.0 \mathrm{~g}\) of water and that the heat capacity for the calorimeter is \(675 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), calculate combustion en\(\operatorname{ergy}(\Delta E)\) for biodiesel in units of \(\mathrm{k} \mathrm{J} / \mathrm{g}\) (b) Calculate combustion energy \((\Delta E)\) for biodiesel in units of \(\mathrm{kJ} / \mathrm{mol}\) (c) If the density of biodiesel is \(0.880 \mathrm{~g} / \mathrm{mL}\), calculate combustion energy \((\Delta E)\) for biodiesel in units of \(\mathrm{kJ} / \mathrm{mL}\).

When \(0.500 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\), is burned in a bomb calorimeter, the temperature of both the water and the calorimeter rise by \(9.15^{\circ} \mathrm{C}\). Assuming that the bath contains \(250.0 \mathrm{~g}\) of water and that the heat capacity for the calorimeter is \(575 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), calculate combustion energies \((\Delta E)\) for ethanol in units of \(\mathrm{kJ} / \mathrm{g}\) and \(\mathrm{kJ} / \mathrm{mol}\)

Under what circumstances are \(\Delta E\) and \(\Delta H\) essentially equal?

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