Problem 98 - Introductory Chemistry Atoms First

Question

Balance each equation: (a) \(\mathrm{Zn}+\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (b) \(\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH}\) (c) \(\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{CS}_{2}+\mathrm{H}_{2}\) (d) \(\mathrm{CO}+\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{O} \mathrm{H}\)

Step-by-Step Solution

Answer

The balanced equations are: (a) \(\mathrm{Zn}+2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (b) \(\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{NaOH}\) (c) \(\mathrm{CH}_{4}+2\mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{CS}_{2}+2\mathrm{H}_{2}\) (d) \(\mathrm{CO}+3\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{O} \mathrm{H}\)

Step 1: Identify unbalanced atoms

In the given equation, there are different numbers of chlorine (Cl) atoms on the reactants and products side. Thus, we need to balance the Cl atoms.

Step 2: Balance Cl atoms

To balance Cl atoms, place a 2 before \(\mathrm{HCl}\) on the reactants side: \[\mathrm{Zn}+2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\] Now the number of Cl atoms is the same on both sides.

Step 3: Check for atom balance

All other atoms (Zn and H) are already balanced. So, the balanced equation is: \[\mathrm{Zn}+2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\] (b) \(\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH}\)

Step 1: Identify unbalanced atoms

In the given equation, there are different numbers of sodium (Na) atoms on the reactants and products sides. We need to balance the Na atoms.

Step 2: Balance Na atoms

To balance Na atoms, place a 2 before \(\mathrm{NaOH}\) on the products side: \[\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{NaOH}\] Now the number of Na atoms is the same on both sides.

Step 3: Check for atom balance

All other atoms (O and H) are already balanced. So, the balanced equation is: \[\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{NaOH}\] (c) \(\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{CS}_{2}+\mathrm{H}_{2}\)

Step 1: Identify unbalanced atoms

In the given equation, there are different numbers of carbon (C) and sulfur (S) atoms on the reactants and products sides. We need to balance these atoms.

Step 2: Balance C and S atoms

To balance C and S atoms, place a 2 before \(\mathrm{H}_{2}S\) on the reactants side and a 2 before \(\mathrm{H}_{2}\) on the products side: \[\mathrm{CH}_{4}+2\mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{CS}_{2}+2\mathrm{H}_{2}\] Now the number of C and S atoms is the same on both sides.

Step 3: Check for atom balance

All other atoms (H) are already balanced. So, the balanced equation is: \[\mathrm{CH}_{4}+2\mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{CS}_{2}+2\mathrm{H}_{2}\] (d) \(\mathrm{CO}+\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{O} \mathrm{H}\)

Step 1: Identify unbalanced atoms

In the given equation, there are different numbers of hydrogen (H) and carbon (C) atoms on the reactants and products sides. We need to balance these atoms.

Step 2: Balance H and C atoms

To balance H and C atoms, place a 3 before \(\mathrm{H}_{2}\) on the reactants side: \[\mathrm{CO}+3\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{O} \mathrm{H}\] Now the number of H and C atoms is the same on both sides.

Step 3: Check for atom balance

All other atoms (O) are already balanced. So, the balanced equation is: \[\mathrm{CO}+3\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{O} \mathrm{H}\]

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in chemical reactions. It allows chemists to determine the correct proportions of substances involved to fully react without excess. To start understanding stoichiometry, imagine it as a recipe: to achieve the desired product, you must mix the correct amounts of ingredients. Each molecule and atom represent a part of the 'recipe' for the chemical reaction.

For example, consider the stoichiometry in the reaction \[ \mathrm{Na}_{2} \mathrm{O} + \mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{NaOH} \]. Here, we see that one mole of sodium oxide (\

Chemical reactions are processes where reactants transform into products. A balanced chemical equation accurately represents the mass and the atoms involved in the reaction respecting the Law of Conservation of Mass. In an equation, reactant and product molecules are depicted using chemical symbols, where each element is denoted by its unique symbol, and compound quantities are shown through coefficients.

For instance, consider the reaction \[ \mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2} + \mathrm{H}_{2} \]. The coefficient '2' in front of HCl indicates there are two moles of hydrochloric acid for every one mole of zinc and one mole of hydrogen gas. When balanced, this equation demonstrates that the same number of atoms for each element are present on both sides of the equation, a fundamental principle of chemical reactions.

Law of Conservation of Mass

The Law of Conservation of Mass is a fundamental principle of science stating that mass cannot be created or destroyed in a closed system through ordinary chemical or physical means. This law is crucial for balancing chemical equations since the mass of reactants must equal the mass of products. During a chemical reaction, atoms are rearranged to form new substances, but the total mass of these atoms remains unchanged.

As demonstrated in the reaction \[ \mathrm{CH}_{4} + 2\mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{CS}_{2} + 2\mathrm{H}_{2} \], the total number of each type of atom is conserved. All carbon, hydrogen, and sulfur atoms are accounted for. No matter how complex a reaction is, the Law of Conservation of Mass guides us to ensure that atoms are neither lost nor gained but simply redistributed.