Balance the following equations using the method outlined in Section 3.7 . (a) \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) (b) \(\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (f) \(\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\) (g) \(\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (h) \(\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}\) (i) \(\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) (j) \(\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (k) \(\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}\) (l) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) \((\mathrm{m}) \mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Step by step solution

01

Balance Equation (a)

For \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\), it is already balanced.

02

Balance Equation (b)

For \(\mathrm{KNO}_{3} \longrightarrow \mathrm{KNO}_{2}+\mathrm{O}_{2}\), you need to balance the oxygen atoms. The balanced equation will then be: \(2KNO_{3}\longrightarrow 2KNO_{2}+O_{2}\).

03

Balance Equation (c)

For \(\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\), you need to equalize the number of nitrogen and hydrogen atoms. The balanced equation is \(NH_{4}NO_{3} \longrightarrow N_{2}O + 2H_{2}O\).

04

Balance Equation (d)

For \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\), the equation is already balanced.

05

Balance Equation (e)

For \(\mathrm{NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\), the balanced equation becomes \(\mathrm{2 NaHCO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) to make the number of Na atoms equal on both sides.

06

Balance Equation (f)

For \(\mathrm{P}_{4} \mathrm{O}_{10}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\), to balance the equation, net equation will be \(P_{4}O_{10}+6H_{2}O \longrightarrow 4H_{3}PO_{4}\).

07

Balance Equation (g)

For \(\mathrm{HCl}+\mathrm{CaCO}_{3} \longrightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\), balancing the equation yields \(2HCl + CaCO_{3} \longrightarrow CaCl_{2} + H_{2}O + CO_{2}\).

08

Balance Equation (h)

In \(\mathrm{Al}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2}\), net equation is \(2Al + 3H_{2}SO_{4} \longrightarrow Al_{2}(SO_{4})_{3}+ 3H_{2}\).

09

Balance Equation (i)

For \(\mathrm{CO}_{2}+\mathrm{KOH} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\), after balancing, the equation becomes: \(CO_{2}+2KOH \longrightarrow K_{2}CO_{3}+ H_{2}O\).

10

Balance Equation (j)

For \(\mathrm{CH}_{4}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\), the balance equation will be \(CH_{4} + 2O_{2} \longrightarrow CO_{2} + 2H_{2}O\).

11

Balance Equation (k)

For \(\mathrm{Be}_{2} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}\), you need to balance the beryllium atoms. Resulting equation is \(Be_{2}C+ 2H_{2}O \longrightarrow 2Be(OH)_{2}+ CH_{4}\).

12

Balance Equation (l)

For \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}(\mathrm{NO}_{3})_{2}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\), we balance the nitrogen atoms to get: \(3Cu + 8HNO_{3} \longrightarrow 3Cu(NO_{3})_{2}+ 2NO + 4H_{2}O\).

13

Balance Equation (m)

To balance \(\mathrm{S}+\mathrm{HNO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\), it needs \(S + 6HNO_{3} \longrightarrow H_{2}SO_{4}+ 4NO_{2} + 2H_{2}O\).

14

Balance Equation (n)

The equation \(\mathrm{NH}_{3}+\mathrm{CuO} \longrightarrow \mathrm{Cu}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) when balanced, will be: \(2NH_{3} + 3CuO \longrightarrow 3Cu + N_{2} + 3H_{2}O\).

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Key Concepts

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

Stoichiometry

Stoichiometry is a branch of chemistry that quantitatively examines the ratios of reactants and products in chemical reactions. In a balanced chemical equation, stoichiometry allows us to predict how much reactant is needed to produce a certain amount of product or vice versa. Think of stoichiometry as a recipe for a chemical equation - it tells us the exact amount of ingredients (reactants) we need to get the final dish (products) we want.

When balancing chemical equations, it is essential to apply the principles of stoichiometry. This means ensuring that the number of atoms of each element involved in the reaction is the same on both sides of the equation. This process respects the law of conservation of mass and allows for the calculation of the stoichiometric coefficients, which are the numbers placed before compounds to indicate the amount (in moles) needed to balance the equation.

For instance, in the balanced reaction of potassium nitrate decomposing into potassium nitrite and oxygen (Step 2), the stoichiometric coefficients are 2, 2 and 1, respectively. This indicates that two moles of potassium nitrate produce two moles of potassium nitrite and one mole of oxygen.

Chemical Reaction

A chemical reaction is a process that results in the transformation of one set of chemical substances to another. This transformation involves breaking and forming bonds between different atoms and molecules. Chemical reactions are represented by chemical equations, which depict the reactants (the starting substances) on the left side and the products (the substances formed) on the right side.

Chemical reactions can be of various types, such as synthesis, decomposition, single replacement, or double replacement, depending on how the atoms reorganize during the process. For example, the decomposition of ammonium nitrate into nitrous oxide and water (Step 3) is a type of reaction where a compound breaks down into multiple products. To fully understand how a reaction proceeds and to be able to perform laboratory experiments correctly, it's important to comprehend the underlying principles involving these transformations and how to balance them correctly.

Law of Conservation of Mass

The law of conservation of mass states that mass—the amount of matter—is neither created nor destroyed in a chemical reaction. This means that the mass of the reactants must equal the mass of the products. When we balance a chemical equation, we are essentially accounting for this law, by ensuring that the number of atoms of each element is the same on both sides of the equation.

Even though atoms rearrange during chemical reactions, the same number of atoms are present from beginning to end. For example, in the reaction where methane burns in oxygen to produce carbon dioxide and water (Step 10), the total mass of methane and oxygen that reacts will equal the total mass of carbon dioxide and water produced. This law is a fundamental concept in chemistry that reinforces the need for balancing chemical equations and allows for predictable quantitative relationships.

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction, showing the reactants and products with their respective quantities. It gives a concise picture of the chemical change involved. Forming a correct chemical equation requires knowledge of the chemical formulas of the substances involved, as well as the ability to balance the equation, which is an application of the law of conservation of mass.

Every chemical equation must be balanced to accurately reflect the conservation of atoms. This requires adjusting coefficients, which are the numbers placed before the chemical formulas, to ensure that for every element involved, the number of atoms on the reactant side is equal to those on the product side. For instance, in balancing the reaction of the formation of beryllium hydroxide from beryllium carbide and water (Step 11), coefficients are adjusted so that the elements beryllium and carbon are balanced. A correctly balanced chemical equation also sets the stage for calculations involving stoichiometry, allowing us to convert between the mass of reactants and products, and to calculate yields and purities for chemical processes.