Question

Balance each of the following equations that describe synthesis reactions. a. \(\mathrm{FeO}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{HCl}(g)\) d. \(\mathrm{K}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{K}_{2} \mathrm{~S}(s)\) e. \(\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Na}_{3} \mathrm{~N}(s)\)

Short answer

The balanced equations are as follows: a. \(4\mathrm{FeO}(s)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\) b. \(2\mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{HCl}(g)\) d. \(2\mathrm{K}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{K}_{2}\mathrm{S}(s)\) e. \(6\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{Na}_{3}\mathrm{N}(s)\)

Step-by-step solution

Identify the starting equation

Our starting equation is: \[\mathrm{FeO}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)\]

Balance Fe and O atoms

To balance the Fe atoms, we need to have the same number of Fe atoms on both sides. Since we have 2 Fe atoms on the product side, we can adjust the coefficient of FeO to 2: \[2\mathrm{FeO}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)\] Now we have 4 O atoms on the reactants side and 3 O atoms on the products side. To balance the O atoms, we can adjust the coefficient of O2 to 1.5 which gives 3 O atoms on each side of the equation: \[2\mathrm{FeO}(s)+1.5\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)\]

Eliminate the fractional coefficient

In order to eliminate the fractional coefficient, we can multiply each coefficient by 2: \[4\mathrm{FeO}(s)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\] The balanced equation for reaction a is: \[4\mathrm{FeO}(s)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\] #b. Balance the CO + O2 equation#

Identify the starting equation

Our starting equation is: \[\mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\]

Balance C and O atoms

The C atoms are already balanced since there's 1 C atom on both sides of the equation. Now we balance the O atoms. Since there's 2 O atoms on the reactants side and only 1 on the products side, we need to change the coefficient of CO2 to 2. We also need to change the coefficient of CO to 2 to balance C again: \[2\mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)\] The balanced equation for reaction b is: \[2\mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)\] #c. Balance the H2 + Cl2 equation#

Identify the starting equation

Our starting equation is: \[\mathrm{H}_{2}(g)+\mathrm{Cl}_{2} (g) \rightarrow \mathrm{HCl}(g)\]

Balance H and Cl atoms

To balance the H atoms, we need to change the coefficient of HCl to 2: \[\mathrm{H}_{2}(g)+\mathrm{Cl}_{2} (g) \rightarrow 2\mathrm{HCl}(g)\] The balanced equation for reaction c is: \[\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{HCl}(g)\] #d. Balance the K + S8 equation#

Identify the starting equation

Our starting equation is: \[\mathrm{K}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{K}_{2}\mathrm{S}(s)\]

Balance K and S atoms

To balance the K atoms, we need to change the coefficient of K to 2: \[2\mathrm{K}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{K}_{2}\mathrm{S}(s)\] The balanced equation for reaction d is: \[2\mathrm{K}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{K}_{2}\mathrm{S}(s)\] #e. Balance the Na + N2 equation#

Identify the starting equation

Our starting equation is: \[\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Na}_{3}\mathrm{N}(s)\]

Balance Na and N atoms

To balance the Na atoms, we need to change the coefficient of Na to 3: \[3\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Na}_{3}\mathrm{N}(s)\] To balance the N atoms, we need to change the coefficient of Na3N to 2: \[3\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{Na}_{3}\mathrm{N}(s)\] To balance Na again, we need to change the coefficient of Na to 6: \[6\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{Na}_{3}\mathrm{N}(s)\] The balanced equation for reaction e is: \[6\mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{Na}_{3}\mathrm{N}(s)\]

Key concepts

Synthesis Reactions

In chemistry, synthesis reactions are a fundamental type as they involve the formation of a compound by combining two or more different reactants. These reactions are often represented in the simple form:

\textbf{A + B → AB}

In the context of the exercise provided, synthesis reactions occur when substances such as metals \textbf{(Fe, K, Na)} combine with nonmetals \textbf{(O, S, N, Cl)} to form complex compounds. For instance, iron oxide is formed by the combination of iron (Fe) and oxygen (O₂), depicted in the balanced chemical equation as:\[\begin{equation}4\text{FeO}(s) + 3\text{O}_2(g) \rightarrow 2\text{Fe}_2\text{O}_3(s)\end{equation}\]
The ability to predict the products of synthesis reactions is highly useful in materials science and in understanding how more complex structures form from simpler ones.

Stoichiometry

Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It serves as the quantitative bridge between the reactants and the products of a chemical equation. The principles of stoichiometry rest on the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

To implement stoichiometry, it is essential to work with a 'balanced' chemical equation, where the number of atoms of each element is the same on both sides of the reaction. For example, in the equation:\[\begin{equation}2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2\end{equation}\]
The stoichiometry shows that 2 moles of carbon monoxide (CO) react with 1 mole of oxygen (O₂) to produce 2 moles of carbon dioxide (CO₂). Understanding stoichiometry is crucial for determining the correct proportions of reactants needed to carry out reactions efficiently in both laboratory and industrial settings.

Chemical Formulas

Chemical formulas represent compounds using the symbols of the elements and numerical subscripts to indicate the number of atoms of each element in a molecule. For example, the chemical formula 'H₂O' indicates that water is composed of two hydrogen atoms and one oxygen atom. The formulas are vital for conveying a great deal of information about the substance they represent, such as the types of atoms present, their proportions, and the overall charge, if any.

Understanding chemical formulas is crucial when balancing equations, as seen in the exercises. Chemical formulas are directly related to the stoichiometry of a reaction since they provide the ratio of atoms in molecules or compounds. Through chemical formulas, one can deduce the quantitative relationships within a chemical equation and thus ensure the conservation of mass and the stoichiometric balance necessary for a chemical reaction to occur as observed in the equation:\[\begin{equation}2\text{K} + \text{S}_8 \rightarrow \text{K}_2\text{S}\end{equation}\]
This chemical formula tells us that potassium sulfide is made from potassium (K) and sulfur (S). The numerical subscript '8' in S₈ indicates that there are eight sulfur atoms present in the reactant molecule.