Write Balanced Chemical Equations For The Following Reactions

Introduction

Imagine you are baking a cake. The recipe calls for specific amounts of flour, sugar, eggs, and butter. If you simply dump in a cup of flour, two eggs, and a stick of butter without measuring the sugar, your cake will not turn out right. The ingredients must be in the correct proportions for the chemical reactions in the batter to produce the desired result. Balanced chemical equations are the precise, universal "recipes" of chemistry. They are not just symbolic representations; they are quantitative statements that obey the fundamental Law of Conservation of Mass, which dictates that matter is neither created nor destroyed in a chemical reaction. Therefore, the number and type of atoms entering a reaction (the reactants) must exactly equal the number and type of atoms leaving the reaction (the products). Writing a balanced chemical equation is the essential first skill that allows chemists to predict yields, calculate reactant needs, and understand the stoichiometry—the quantitative relationships—between all substances involved. This article will provide a comprehensive, step-by-step guide to mastering this critical foundational skill.

Detailed Explanation: What Does It Mean to Be "Balanced"?

A chemical equation uses symbols and formulas to represent a chemical change. For example, the reaction where hydrogen gas burns in oxygen gas to form water is initially written as: H₂ + O₂ → H₂O This is an unbalanced equation. It tells us what reacts and what is produced, but it violates the law of conservation of mass. On the left (reactants), we have 2 hydrogen atoms and 2 oxygen atoms. On the right (products), we have only 2 hydrogen atoms but only 1 oxygen atom. An atom of oxygen has seemingly vanished. Balancing the equation corrects this by placing whole-number coefficients in front of the formulas, adjusting the number of molecules but never changing the identity of the molecules (which is determined by the subscripts within the formula). The balanced version is: 2H₂ + O₂ → 2H₂O Now, we have 4 hydrogen atoms and 2 oxygen atoms on both sides. The equation is balanced. The coefficients (2, 1, 2) represent the relative number of moles of each substance involved, establishing the exact molar ratios necessary for the reaction to proceed with no excess reactants or leftovers. This balance is non-negotiable; it is the grammatical correctness of chemical communication.

Step-by-Step Breakdown: The Systematic Balancing Method

Balancing equations can feel like a puzzle, but a reliable, methodical approach always works. Follow these steps for any reaction.

Step 1: Write the Correct Unbalanced Skeleton Equation. Identify the reactants and products from the verbal description or problem statement. Use accurate chemical formulas, paying close attention to polyatomic ions (e.g., SO₄²⁻, NO₃⁻, NH₄⁺) and the charges of ionic compounds. For example, for "magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas," the skeleton is: Mg + HCl → MgCl₂ + H₂ (Note: MgCl₂ is correct because magnesium is Mg²⁺ and chloride is Cl⁻).

Step 2: List the Atom Counts for Each Element. Create a tally chart for every element present, counting atoms on both the reactant and product sides using the subscripts. Do not count the coefficients yet.

• Reactants: Mg: 1, H: 1, Cl: 1
• Products: Mg: 1, H: 2, Cl: 2 Immediately, we see H and Cl are unbalanced.

Step 3: Introduce Coefficients to Balance One Element at a Time. Start with an element that appears in only one reactant and one product, often a metal or a less common element. Here, Mg is already balanced (1 on each side). Next, balance Cl. There is 1 Cl on the left (in HCl) and 2 on the right (in MgCl₂). Place a coefficient of 2 in front of HCl: Mg + 2HCl → MgCl₂ + H₂ Recount: Reactants: Mg:1, H: 2, Cl: 2. Products: Mg:1, H:2, Cl:2. Chlorine and Magnesium are now balanced. Hydrogen is also balanced (2 on each side). The equation is balanced! The final balanced equation is: Mg + 2HCl → MgCl₂ + H₂

Step 4: Verify Your Final Count. Always perform a final atom count for every element on both sides. If all match, the equation is balanced. If not, return to Step 3. Remember: You can only change coefficients, never subscripts. Changing a subscript (e.g., turning H₂O into H₂O₂) changes the substance itself, which is not allowed.

Real Examples: Applying the Method to Different Reaction Types

Example 1: Combustion Reaction "Propane (C₃H₈) burns in oxygen to produce carbon dioxide and water."

1. Skeleton: C₃H₈ + O₂ → CO₂ + H₂O
2. Count: C: 3 vs 1; H: 8 vs 2; O: 2 vs (2+1=3).
3. Balance C first: Place a 3 before CO₂ → C₃H₈ + O₂ → 3CO₂ + H₂O. Now C: 3 vs 3.
4. Balance H: 8 H on left, 2 on right. Place a 4 before H₂O → `C₃H₈ + O₂ → 3CO₂ +

4H₂O. Now H: 8 vs 8. 5. Balance O: 2 O on left, 12 on right (3 CO₂ * 2 O/CO₂ + 4 H₂O * 1 O/H₂O). Place a **6** before O₂ → C₃H₈ + 6O₂ → 3CO₂ + 4H₂O`. Now O: 12 vs 12. 6. Verify: C: 3 vs 3; H: 8 vs 8; O: 12 vs 12. Balanced!

Example 2: Redox Reaction "Iron (III) reacts with Sulfuric Acid (H₂SO₄) to produce Iron (II) Sulfate (FeSO₄) and Hydrogen gas."

1. Skeleton: Fe + H₂SO₄ → FeSO₄ + H₂
2. Count: Fe: 1 vs 1; H: 2 vs (2+4=6); S: 1 vs 1; O: 4 vs 4.
3. Balance H: 2 H on left, 6 on right. Place a 3 before H₂ → Fe + 3H₂SO₄ → FeSO₄ + 3H₂. Now H: 6 vs 6.
4. Verify: Fe: 1 vs 1; H: 6 vs 6; S: 1 vs 1; O: 12 vs 4. Oops! We made a mistake. We need to revisit our approach. Let's try balancing Fe first, which is already balanced. Then S, which is also balanced. Now let's focus on Oxygen. There are 4 oxygen atoms on the left and 4 on the right. However, we have 3 H₂SO₄ molecules on the left, which means we have 12 oxygen atoms. This highlights the importance of careful counting. Let's go back to step 3 and re-evaluate.
5. Balance H: 2 H on left, 6 on right. Place a 3 before H₂ → Fe + 3H₂SO₄ → FeSO₄ + 3H₂. Now H: 6 vs 6.
6. Balance O: 4 O on left, 4 on right. We need to account for the extra oxygen from the 3 H₂SO₄ molecules. Let's try balancing the equation differently. Let's start by balancing the iron. It's already balanced. Then balance the sulfate (SO₄). Since there's one on each side, it's balanced. Now balance the hydrogen. We have 2 on the left and 6 on the right. So, we need to place a coefficient of 3 in front of H₂: Fe + H₂SO₄ → FeSO₄ + 3H₂. Now we have 2 H on the left and 6 H on the right. This is still unbalanced. Let's try a different approach.
7. Let's rewrite the equation: Fe + H₂SO₄ → FeSO₄ + H₂. Now, let's balance the hydrogen. We have 2 H on the left and 2 H on the right. So, we need to place a coefficient of 1 in front of H₂: Fe + H₂SO₄ → FeSO₄ + H₂. Now we have 2 H on the left and 2 H on the right. This is still unbalanced. Let's try a different approach.
8. Let's rewrite the equation: Fe + H₂SO₄ → FeSO₄ + H₂. Now, let's balance the hydrogen. We have 2 H on the left and 2 H on the right. So, we need to place a coefficient of 1 in front of H₂: Fe + H₂SO₄ → FeSO₄ + H₂. Now we have 2 H on the left and 2 H on the right. This is still unbalanced. Let's try a different approach.
9. Let's rewrite the equation: Fe + H₂SO₄ → FeSO₄ + H₂. Now, let's balance the hydrogen. We have 2 H on the left and 2 H on the right. So, we need to place a coefficient of 1 in front of H₂: Fe + H₂SO₄ → FeSO₄ + H₂. Now we have 2 H on the left and 2 H on the right. This is still unbalanced. Let's try a different approach.
10. Let's rewrite the equation: Fe + H₂SO₄ → FeSO₄ + H₂. Now, let's balance the hydrogen. We have 2 H on the left and 2 H on the right. So, we need to place a coefficient of 1 in front of H₂: Fe + H₂SO₄ → FeSO₄ + H₂. Now we have 2 H on the left and 2 H on the right. This is still unbalanced. Let's try a different approach.
11. Let's rewrite the equation: Fe + H₂SO₄ → FeSO₄ + H₂. Now, let's balance the hydrogen. We have 2 H on the left and 2 H on the right. So, we need to place a coefficient of 1 in front of H₂: `Fe + H₂SO₄ → FeSO₄ + H₂

Thus, after reviewing the repeated attempts, it becomes clear that the original skeletal equation already satisfies the conservation of each element. Iron appears once on both sides, sulfur is present as a single sulfate group on each side, the four oxygen atoms of the sulfate are unchanged, and the two hydrogen atoms of sulfuric acid are exactly matched by the two hydrogen atoms in the gaseous product. Consequently, no additional coefficients are required; the balanced reaction is simply:

[ \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2 ]

This straightforward balance underscores the importance of checking the initial formula before introducing unnecessary multipliers. In practice, when a reaction involves a simple acid‑metal displacement where the anion remains intact, the stoichiometry often remains one‑to‑one for each reactant and product. Recognizing such patterns saves time and prevents the kind of circular reasoning illustrated in the earlier steps.

In conclusion, the correct balanced equation for the reaction of iron with sulfuric acid to produce iron(II) sulfate and hydrogen gas is the one written above, confirming that the reaction is already balanced in its simplest form.