Question

Balance each of the following chemical equations. a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q) \rightarrow \mathrm{BaCrO}_{4}(s)+\mathrm{NaNO}_{3}(a q)\) b. \(\mathrm{PbCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{KCl}(a q)\) c. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) e. \(\operatorname{Sr}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) f. \(\mathrm{BaO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) g. \(\operatorname{AsI}_{3}(s) \rightarrow \operatorname{As}(s)+\mathrm{I}_{2}(s)\) h. \(\mathrm{CuSO}_{4}(a q)+\mathrm{KI}(s) \rightarrow \mathrm{CuI}(s)+\mathrm{I}_{2}(s)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)\)

Short answer

The balanced chemical equations are as follows: a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2}\mathrm{CrO}_{4}(a q) \rightarrow \mathrm{BaCrO}_{4}(s)+2\mathrm{NaNO}_{3}(a q)\) b. \(\mathrm{PbCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2\mathrm{KCl}(a q)\) c. \(\mathrm{C}_{2}\mathrm{H}_{5} \mathrm{OH}(l) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2}\mathrm{O}(l)\) d. \(\mathrm{CaC}_{2}(s) + 2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) + \mathrm{C}_{2} \mathrm{H}_{2}(g)\) e. \(\operatorname{Sr}(s) + 2\mathrm{HNO}_{3}(a q) \rightarrow \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) f. \(\mathrm{BaO}_{2}(s) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s) + 2\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) g. \(2\operatorname{AsI}_{3}(s) \rightarrow 2\operatorname{As}(s) + 3\mathrm{I}_{2}(s)\) h. \(\mathrm{CuSO}_{4}(a q) + 2\mathrm{KI}(s) \rightarrow \mathrm{CuI}(s) + \mathrm{I}_{2}(s) + \mathrm{K}_{2} \mathrm{SO}_{4}(a q)\)

Step-by-step solution

a. Balancing the equation for the reaction of Barium nitrate and Sodium chromate

Balance the given chemical equation: \[ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q) \rightarrow \mathrm{BaCrO}_{4}(s)+\mathrm{NaNO}_{3}(a q) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q) \rightarrow \mathrm{BaCrO}_{4}(s)+2\mathrm{NaNO}_{3}(a q) \]

b. Balancing the equation for the reaction of Lead(II) chloride and Potassium sulfate

Balance the given chemical equation: \[ \mathrm{PbCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+\mathrm{KCl}(a q) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \mathrm{PbCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2\mathrm{KCl}(a q) \]

c. Balancing the equation for the combustion of Ethanol

Balance the given chemical equation: \[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \mathrm{C}_{2}\mathrm{H}_{5} \mathrm{OH}(l) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2}\mathrm{O}(l) \]

d. Balancing the equation for the reaction of Calcium carbide and Water

Balance the given chemical equation: \[ \mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \mathrm{CaC}_{2}(s) + 2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) + \mathrm{C}_{2} \mathrm{H}_{2}(g) \]

e. Balancing the equation for the reaction of Strontium and Nitric acid

Balance the given chemical equation: \[ \operatorname{Sr}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}(g) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \operatorname{Sr}(s) + 2\mathrm{HNO}_{3}(a q) \rightarrow \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}(g) \]

f. Balancing the equation for the reaction of Barium peroxide and Sulfuric acid

Balance the given chemical equation: \[ \mathrm{BaO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \mathrm{BaO}_{2}(s) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s) + 2\mathrm{H}_{2} \mathrm{O}_{2}(a q) \]

g. Balancing the equation for the decomposition of Arsenic(III) iodide

Balance the given chemical equation: \[ \operatorname{AsI}_{3}(s) \rightarrow \operatorname{As}(s)+\mathrm{I}_{2}(s) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ 2\operatorname{AsI}_{3}(s) \rightarrow 2\operatorname{As}(s) + 3\mathrm{I}_{2}(s) \]

h. Balancing the equation for the reaction of Copper(II) sulfate and Potassium iodide

Balance the given chemical equation: \[ \mathrm{CuSO}_{4}(a q)+\mathrm{KI}(s) \rightarrow \mathrm{CuI}(s)+\mathrm{I}_{2}(s)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \] Step 1: Start by counting the number of atoms for each type in both the reactants and the products. Step 2: Adjust the coefficients until the number of atoms of each type are equal on both sides of the equation. Step 3: The balanced equation is: \[ \mathrm{CuSO}_{4}(a q) + 2\mathrm{KI}(s) \rightarrow \mathrm{CuI}(s) + \mathrm{I}_{2}(s) + \mathrm{K}_{2} \mathrm{SO}_{4}(a q) \]

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is like the cookbook of chemistry, where you need to follow a recipe, not only including the ingredients but also their correct proportions. This ensures that all the molecules are balanced properly and none of them are wasted.
To get started with stoichiometry, you first need a balanced chemical equation, as this tells you the proportion of each substance involved. This is where balancing equations comes into play. For example, if you're baking a cake and the recipe calls for one egg and two cups of flour, you can't switch them around. While in chemical reactions, stoichiometry ensures that you'll have the right amount of each reactant turning into products, avoiding excess waste.

In practice, stoichiometry involves:

• Counting the number of atoms for each element on both sides of the chemical equation.
• Making sure the equation is balanced by adjusting the coefficients placed in front of compounds or elements.

This keeps the law of conservation of mass in check, where matter is neither created nor destroyed in a reaction. The balanced equation then acts as a guide to calculate how much of each reactant is needed and how much product will be formed.

Reaction Types

Chemical reactions can be classified into different types based on how reactants are transformed into products. Understanding the type of reaction can help you anticipate the changes that will occur during the process.

Here are some common types of chemical reactions:

• **Synthesis reactions**: Two or more substances combine to form a more complex compound. For example, the reaction between iron and sulfur to create iron sulfide.
• **Decomposition reactions**: A single compound breaks down into two or more simpler substances. This is what occurs in the decomposition equation of arsenic(III) iodide to arsenic and iodine gas.
• **Single replacement reactions**: One element replaces another in a compound. An example is the reaction of strontium with nitric acid, where strontium replaces hydrogen.
• **Double replacement reactions**: The ions in two compounds swap partners, leading to the formation of two new compounds. The reaction of barium nitrate with sodium chromate is an example.
• **Combustion reactions**: These involve oxygen and produce energy in the form of heat or light. The burning of ethanol is a classic combustion reaction.

Recognizing these patterns helps in predicting reaction products and understanding how to balance equations effectively.

Chemical Reactions

A chemical reaction involves changing substances or reactants into new substances or products. This transformation happens at the molecular level, involving the breaking and forming of bonds. During a chemical reaction, the identity and properties of the reactants change to produce different substances, with different properties.

Indicators of a chemical reaction include:

• Color changes, indicating new substances are forming.
• Temperature changes, often exothermic (releasing heat) or endothermic (absorbing heat).
• Gas production, often visible as bubbles.
• Precipitate formation, where a solid forms from a liquid solution.

These markers help you to determine if a chemical reaction has occurred. Each reaction is unique and has its own set of reactants and products. Balancing chemical equations is an essential skill in chemistry as it ensures that the number of atoms for each element is conserved in a reaction.
Understanding chemical reactions also involves knowing that they follow specific laws, such as the law of conservation of mass. This law states that atoms are neither created nor destroyed in a chemical reaction, necessitating the balancing of chemical equations. Knowing this principle reinforces the importance of properly balancing equations in stoichiometry.