Balancing Chemical Equations 🧮 Balancing chemical equations is a crucial step in chemistry. The goal is to ensure that the number of atoms of each element is equal on both the reactant and product sides of the equation. Combustion Reaction 🔥 Let's consider a combustion reaction: C₃H₈ + O₂ → CO₂ + H₂O To balance this equation, we need to ensure that the number of atoms of each element is equal on both sides. Carbon Atoms We have 3 carbon atoms on the left side (C₃H₈). We need 3 carbon atoms on the right side. To achieve this, we put a 3 in front of CO₂. Hydrogen Atoms We have 8 hydrogen atoms on the left side (C₃H₈). We need 8 hydrogen atoms on the right side. To achieve this, we put a 4 in front of H₂O. Oxygen Atoms We have 10 oxygen atoms on the right side (6 in CO₂ and 4 in H₂O). We need 10 oxygen atoms on the left side. To achieve this, we put a 5 in front of O₂. Balancing Another Equation Let's consider another equation: C₄H₁₀ + O₂ → CO₂ + H₂O To balance this equation, we need to follow the same steps as before. Carbon Atoms We have 4 carbon atoms on the left side (C₄H₁₀). We need 4 carbon atoms on the right side. To achieve this, we put a 4 in front of CO₂. Hydrogen Atoms We have 10 hydrogen atoms on the left side (C₄H₁₀). We need 10 hydrogen atoms on the right side. To achieve this, we put a 5 in front of H₂O. Oxygen Atoms We have 13 oxygen atoms on the right side (8 in CO₂ and 5 in H₂O). We need 13 oxygen atoms on the left side. To achieve this, we need to multiply the equation by 2. Reactant Coefficient C₄H₁₀ 2 O₂ 13 Product Coefficient CO₂ 8 H₂O 10 Even-Odd Situation When we encounter an even-odd situation, we can rectify it by multiplying the equation by a suitable factor. Definition: An even-odd situation occurs when we have an even number of atoms of an element on one side of the equation and an odd number of atoms on the other side. For example, consider the equation: Al + HCl → AlCl₃ + H₂ To balance this equation, we need to put a 6 in front of HCl to get 6 chlorine atoms on both sides. However, this would give us an odd number of hydrogen atoms on the left side and an even number on the right side. To rectify this, we can multiply the equation by 2. Balancing Another Equation Let's consider another equation: Ga + CuBr₂ → GaBr₃ + Cu To balance this equation, we need to follow the same steps as before. Bromine Atoms We have 2 bromine atoms on the left side (CuBr₂). We need 3 bromine atoms on the right side (GaBr₃). To achieve this, we need to put a 3 in front of CuBr₂. Copper Atoms We have 1 copper atom on the left side (CuBr₂). We need 1 copper atom on the right side (Cu). To achieve this, we need to put a 1 in front of Cu. Gallium Atoms We have 1 gallium atom on the left side (Ga). We need 1 gallium atom on the right side (GaBr₃). To achieve this, we need to put a 1 in front of Ga. Reactant Coefficient Ga 2 CuBr₂ 3 Product Coefficient GaBr₃ 2 Cu 3 More Examples Let's consider more examples: I₂ + F₂ → IF₇ SO₂ + O₂ → SO₃ Na + S₈ → Na₂S Na₃PO₄ + MgCl₂ → NaCl + Mg₃(PO₄)₂ To balance these equations, we need to follow the same steps as before. I₂ + F₂ → IF₇ We have 2 iodine atoms on the left side (I₂). We need 2 iodine atoms on the right side (IF₇). To achieve this, we need to put a 2 in front of I₂. SO₂ + O₂ → SO₃ We have 4 oxygen atoms on the left side (2 in SO₂ and 2 in O₂). We need 3 oxygen atoms on the right side (SO₃). To achieve this, we need to multiply the equation by 2. Na + S₈ → Na₂S We have 8 sulfur atoms on the left side (S₈). We need 1 sulfur atom on the right side (Na₂S). To achieve this, we need to put a 16 in front of Na. Na₃PO₄ + MgCl₂ → NaCl + Mg₃(PO₄)₂ We have 1 phosphate unit on the left side (Na₃PO₄). We need 2 phosphate units on the right side (Mg₃(PO₄)₂). To achieve this, we need to put a 2 in front of Na₃PO₄. By following these steps, we can balance any chemical equation. Balancing Chemical Equations 🔩 Balancing Reaction: Aluminum Chloride and Potassium Chloride To balance the reaction: AlCl₃ + KCl → KAlCl₄ We have 2 aluminum atoms on both sides. On the left side, we have 6 chlorine atoms, so we need to put a 6 in front of KCl. On the right side, we have 3 times 2 = 6 potassium atoms. Everything is balanced at this point. Coefficients: 3, 2, 6, and 1. Balancing Reaction: Ammonia and Oxygen Gas To balance the reaction: NH₃ + O₂ → NO + H₂O We have: 1 nitrogen atom on both sides (balanced) 3 hydrogen atoms on the left side, 2 on the right side (not balanced) Step 1: Balance Hydrogen Atoms To balance the hydrogen atoms, we double the number of hydrogen atoms on the left side: 3 × 2 = 6. We need 3 in front of H₂O. Step 2: Balance Nitrogen Atoms We have 2 nitrogen atoms on the left side, 1 on the right side. We need to put a 2 in front of NO. Step 3: Balance Oxygen Atoms We have 5 oxygen atoms in total (2 in NO, 3 in H₂O). We need to multiply everything by 2 to eliminate the fraction: 2NH₃ + 5O₂ → 2NO + 3H₂O Coefficients: 2, 5, 2, and 3. Balancing Reaction: Ethanol and Oxygen Gas To balance the reaction: C₂H₅OH + O₂ → CO₂ + H₂O Step 1: Balance Carbon Atoms We have 2 carbon atoms on the left side, so we need to put a 2 in front of CO₂. Step 2: Balance Hydrogen Atoms We have 6 hydrogen atoms on the left side (5 in C₂H₅OH, 1 in H₂O). We need 3 water molecules to balance the hydrogen atoms. Step 3: Balance Oxygen Atoms We have 4 oxygen atoms in the 2CO₂ molecules, 3 oxygen atoms in the 3H₂O molecules, and 1 oxygen atom in C₂H₅OH. We need a total of 7 oxygen atoms on both sides. Atom Left Side Right Side Carbon 2 2 Hydrogen 6 6 Oxygen 7 7 Let's call the number of oxygen atoms from O₂ "x". We need to solve the equation: 1 + x = 4 + 3. Solving for x, we get x = 6. We need 6 oxygen atoms from O₂, which is 3 times 2. So, we put a 3 in front of O₂.