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Balance the following chemical equations. \(\mathrm{MnO}_{2}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2} \mathrm{O}_{3}(a q)+\mathrm{CO}_{2}(g)\) \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{I}_{2}(g)\) \(\mathrm{KOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
The balanced chemical equations are: 1. \(2\mathrm{MnO}_{2}(s) + 4\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(a q) + 2\mathrm{CO}_{2}(g)\) 2. \(2\mathrm{Al}(s) + 3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + 3\mathrm{H}_{2}(g)\) 3. \(\mathrm{C}_{4}\mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l)\) 4. \(2\mathrm{NH}_{4} \mathrm{I}(a q) + \mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{NH}_{4} \mathrm{Cl}(a q) + \mathrm{I}_{2}(g)\) 5. \(2\mathrm{KOH}(a q) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\)

Step by step solution

01

Balancing Equation 1

\(\mathrm{MnO}_{2}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(a q)+\mathrm{CO}_{2}(g)\) 1. First, balance the manganese (\(\text{Mn}\)) atoms: \(2 \times \mathrm{MnO}_{2}\rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}\) 2. Then, balance the oxygen (\(\text{O}\)) atoms: \(4 \times \mathrm{CO}\rightarrow 2 \times \mathrm{CO}_{2}\) The balanced equation is: \(2\mathrm{MnO}_{2}(s) + 4\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(a q) + 2\mathrm{CO}_{2}(g)\)
02

Balancing Equation 2

\(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) 1. Start with balancing the aluminium (\(\text{Al}\)) atoms: \(2 \times \mathrm{Al} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) 2. Then, balance the sulfate (\(\mathrm{SO}_{4}\)) ions: \(3 \times \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) 3. Finally, balance the hydrogen (\(\text{H}\)) atoms: \(3 \times \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 6 \times \mathrm{H}_{2}\) The balanced equation is: \(2\mathrm{Al}(s) + 3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + 3\mathrm{H}_{2}(g)$
03

Balancing Equation 3

\(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) 1. Start by balancing the carbon (\(\text{C}\)) atoms: \(\mathrm{C}_{4}\mathrm{H}_{10} \rightarrow 4 \times \mathrm{CO}_{2}\) 2. Then, balance the hydrogen (\(\text{H}\)) atoms: \(\mathrm{C}_{4}\mathrm{H}_{10} \rightarrow 5 \times \mathrm{H}_{2} \mathrm{O}\) 3. Finally, balance the oxygen (\(\text{O}\)) atoms: \(13/2 \times \mathrm{O}_{2} \rightarrow 4 \times \mathrm{CO}_{2} + 5 \times \mathrm{H}_{2} \mathrm{O}\) The balanced equation is: \(\mathrm{C}_{4}\mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l)\)
04

Balancing Equation 4

\(\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{I}_{2}(g)\) 1. First, balance the iodine (\(\text{I}\)) atoms: \(2 \times \mathrm{NH}_{4}\mathrm{I} \rightarrow \mathrm{I}_{2}\) 2. Then, balance the chlorine (\(\text{Cl}\)) atoms: \(\mathrm{NH}_{4}\mathrm{I} + \mathrm{Cl}_{2} \rightarrow 2 \times \mathrm{NH}_{4}\mathrm{Cl}\) The balanced equation is: \(2\mathrm{NH}_{4} \mathrm{I}(a q) + \mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{NH}_{4} \mathrm{Cl}(a q) + \mathrm{I}_{2}(g)\)
05

Balancing Equation 5

\(\mathrm{KOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) 1. First, balance the potassium (\(\text{K}\)) atoms: \(2 \times \mathrm{KOH} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4}\) 2. Then, balance the sulfate (\(\mathrm{SO}_{4}\)) ions: \(\mathrm{KOH}+ \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4}\) 3. Finally, balance the water (\(\mathrm{H}_{2}\mathrm{O}\)) molecules: \(2 \times \mathrm{KOH} \rightarrow 2 \times \mathrm{H}_{2}\mathrm{O}\) The balanced equation is: \(2\mathrm{KOH}(a q) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\)

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Key Concepts

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

Stoichiometry
Stoichiometry can be thought of as the 'mathematics' behind chemistry. It provides quantitative relationships between substances involved in a chemical reaction. In practice, stoichiometry is used to calculate the amount of reactants required to produce a desired amount of product. A stoichiometric coefficient represents the number of molecules or moles of a substance involved in a reaction.

Let's consider an example: in the chemical equation \(2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}\), the stoichiometric coefficients are 2 for hydrogen, 1 for oxygen, and 2 for water. They indicate that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

When balancing equations, it’s crucial to adjust these coefficients to respect the conservation of mass, ensuring that the same number of atoms for each element is present on both sides of the equation. Understanding stoichiometry is not only fundamental for balancing chemical equations but also essential for empirical calculations, determining reaction yields, and performing conversions such as moles to grams, and vice versa.
Chemical Reaction
A chemical reaction is a process in which substances, known as reactants, transform into new substances called products. During this process, the bonds between atoms are broken and new ones are formed. There are different types of chemical reactions, including combination, decomposition, single-replacement, double-replacement, and combustion.

Each chemical reaction is represented by a chemical equation which shows the symbols and formulas of the reactants and products. For instance, in the combustion reaction \(\mathrm{C}_4\mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_2(g) + 5\mathrm{H}_2 \mathrm{O}(l)\), butane gas (\text{C}_4\text{H}_{10}) reacts with oxygen (\text{O}_2) to produce carbon dioxide (\text{CO}_2) and water (\text{H}_2\text{O}).

For a chemical reaction to be useful and accurate, the equation representing it must be balanced so that the same number of each type of atom appears on both the reactant and product sides. This principle is known as the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.
Mole Concept
The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of particles it contains. A mole is defined as exactly 6.02214076 x 10^23 particles (Avogadro's number), which could be atoms, molecules, ions, or electrons depending on the context.

For example, one mole of carbon atoms weighs 12 grams, and because of Avogadro's number, it contains exactly 6.022 x 10^23 carbon atoms. The mole concept allows chemists to count atoms and molecules by weighing, which is a pragmatic approach to handling vast numbers of atomic or molecular entities.

Understanding the mole concept is crucial in stoichiometry for conversions between the mass of a substance and the amount in moles. It is critical to recognize that when we balance a chemical equation, we are actually balancing the moles of the reactants and products according to the ratios provided by the balanced equation, ensuring that mass and the number of particles are conserved.

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Most popular questions from this chapter

A common experiment in introductory chemistry courses involves heating a weighed mixture of potassium chlorate, \(\mathrm{KClO}_{3}\), and potassium chloride. Potassium chlorate decomposes when heated, producing potassium chloride and evolving oxygen gas. By measuring the volume of oxygen gas produced in this experiment, students can calculate the relative percentage of \(\mathrm{KClO}_{3}\) and \(\mathrm{KCl}\) in the original mixture. Write the balanced chemical equation for this process.

Nitrous oxide gas (systematic name: dinitrogen monoxide) is used by some dental practitioners as an anesthetic. Nitrous oxide (and water vapor as by- product) can be produced in small quantities in the laboratory by careful heating of ammonium nitrate. Write the unbalanced chemical equation for this reaction.

When iron wire is heated in the presence of sulfur, the iron soon begins to glow, and a chunky, blue-black mass of iron(II) sulfide is formed. Write the unbalanced chemical equation for this reaction.

When solid red phosphorus, \(\mathrm{P}_{4},\) is burned in air, the phosphorus combines with oxygen, producing a choking cloud of tetraphosphorus decoxide. Write the unbalanced chemical equation for this reaction.

Which of the following statements about chemical equations is (are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right, and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed, so a chemical equation must have the same number of atoms on both sides of the equation.

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