Question

Balance each of the following equations that describe synthesis reactions. a. \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)\) b. \(\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{P}_{2} \mathrm{O}_{5}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)\)

Short answer

a. \(CaO(s) + 2H_2O(l) \rightarrow Ca(OH)_2(s)\) b. \(4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)\) c. \(P_2O_5(s) + 3H_2O(l) \rightarrow 2H_3PO_4(aq)\)

Step-by-step solution

Balance equation a: CaO(s) + H₂O(l) → Ca(OH)₂(s)

To balance this equation, note that there is already one calcium atom on each side. However, the number of hydrogen and oxygen atoms is not equal on both sides. There is one oxygen atom on the reactant side and two oxygen atoms on the product side. Additionally, there are two hydrogen atoms on the reactant side but four hydrogen atoms on the product side. To balance the equation, add a coefficient in front of the H₂O to make the number of hydrogen and oxygen atoms equal on both sides: \(CaO(s) + 2H_2O(l) \rightarrow Ca(OH)_2(s)\) Now the equation is balanced, as there are two oxygen and four hydrogen atoms on both sides.

Balance equation b: Fe(s) + O₂(g) → Fe₂O₃(s)

To balance this equation, start by looking at the number of iron atoms. There are two iron atoms on the product side, but only one on the reactant side. Therefore, we need to add a coefficient in front of the Fe to balance the iron atoms: \(2Fe(s) + O_2(g) \rightarrow Fe_2O_3(s)\) Now the number of iron atoms is balanced on both sides of the equation. However, the number of oxygen atoms is still unbalanced. There are three oxygen atoms on the product side and only two on the reactant side. To balance the number of oxygen atoms, add a coefficient in front of the O₂ to get: \(2Fe(s) + \frac{3}{2} O_2(g) \rightarrow Fe_2O_3(s)\) It's better to avoid using fractional coefficients in balanced chemical equations, so multiply the entire equation by 2 to get whole number coefficients: \(4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)\) Now the equation is balanced, with four iron atoms and six oxygen atoms on both sides.

Balance equation c: P₂O₅(s) + H₂O(l) → H₃PO₄(aq)

First, note that there are two phosphorus atoms on the reactant side and only one on the product side. Add a coefficient in front of H₃PO₄ to balance the phosphorus atoms: \(P_2O_5(s) + H_2O(l) \rightarrow 2H_3PO_4(aq)\) Next, notice that there are five oxygen atoms on the reactant side and eight on the product side. There are also two hydrogen atoms on the reactant side and twelve hydrogen atoms on the product side. To balance the oxygen and hydrogen atoms, put a coefficient in front of H₂O: \(P_2O_5(s) + 3H_2O(l) \rightarrow 2H_3PO_4(aq)\) Now the equation is balanced, with two phosphorus atoms, eight oxygen atoms, and twelve hydrogen atoms on both sides.

Key concepts

Synthesis Reactions

Synthesis reactions are a fundamental type of chemical reaction where two or more simple substances combine to form a more complex compound. These reactions can be easily identified by their general form, which is \( A + B \rightarrow AB \). In the given exercise examples, we encounter various synthesis reactions, each involving the combination of elements or compounds to form a new product.
For instance, in the reaction where calcium oxide reacts with water to form calcium hydroxide (\(CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(s)\)), we can see that a solid and a liquid combine to create a new solid substance. Similarly, iron combining with oxygen to form iron(III) oxide (\(Fe(s) + O_2(g) \rightarrow Fe_2O_3(s)\)) is another example of a synthesis reaction. It is crucial for students to recognize the pattern of reactants combining and the increase in complexity in the product to identify synthesis reactions effectively.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It involves using the coefficients of the substances in a balanced chemical equation to determine the relative amounts needed for the reaction to occur. This process is vital to ensure that the law of conservation of mass is obeyed, meaning that matter is neither created nor destroyed in the reaction.

In the balancing of the equations provided in the exercise, stoichiometry is demonstrated by adjusting coefficients to ensure that the number of atoms of each element is the same on both sides of the equation. For example, when balancing the reaction \(4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)\), stoichiometry guides us to use coefficients like 4 for iron and 3 for oxygen to satisfy the stoichiometric conditions. This highlights the stoichiometric ratio of 4:3:2 for iron, oxygen, and iron(III) oxide, respectively. Understanding stoichiometry is crucial for students as it lays the groundwork for further studies in chemistry, such as calculations involving reactant and product masses, gas volumes, and solution concentrations.

Chemical Reactions

Chemical reactions are processes where reactants transform into products through the breaking and forming of chemical bonds. The reaction types vary widely, from synthesis reactions to decomposition, combustion, single replacement, and double replacement reactions, among others. Each type of reaction has its distinctive characteristics and reaction patterns.

An essential aspect of studying chemical reactions is the ability to write and balance chemical equations accurately. The balanced equations reflect the conservation of mass and charge, which are fundamental principles in chemistry. When approaching the balance of any chemical reaction, like \(P_2O_5(s) + 3H_2O(l) \rightarrow 2H_3PO_4(aq)\), one must first identify the reactants and products, then use stoichiometry to ensure that there are equal numbers of atoms of each element on both sides of the equation. Through practice and understanding of different reaction types, students can develop a stronger grasp of how to predict the products of a reaction and the reactants required, which is a crucial skill in the field of chemistry.