The balanced chemical equation is

← Back to Questions

### Correct Answer The balanced chemical equation is: \(\boldsymbol{CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl}\) ### Detailed Steps for Solving the Problem To balance the equation \(\ce{CH_4 + Cl_2 \rightarrow CCl_4 + HCl}\), follow these steps: #### Step 1: Identify Atoms and Their Counts List the number of each atom on the **reactant side** (left) and **product side** (right): | Atom | Reactant Side (\(\ce{CH_4 + Cl_2}\)) | Product Side (\(\ce{CCl_4 + HCl}\)) | |------|---------------------------------------|--------------------------------------| | C | 1 (from \(\ce{CH_4}\)) | 1 (from \(\ce{CCl_4}\)) | | H | 4 (from \(\ce{CH_4}\)) | 1 (from \(\ce{HCl}\)) *per molecule* | | Cl | 2 (from \(\ce{Cl_2}\)) *per molecule* | 4 (from \(\ce{CCl_4}\)) + 1 (from \(\ce{HCl}\)) *per molecule* | #### Step 2: Balance Atoms One by One We use **coefficients** (numbers in front of formulas) to adjust atom counts. Coefficients multiply all atoms in the formula they precede (e.g., \(4\ce{Cl_2}\) means 4 molecules of \(\ce{Cl_2}\), containing \(4 \times 2 = 8\) Cl atoms). - **Carbon (C):** Already balanced (1 on both sides). - **Hydrogen (H):** Reactant side has 4 H (from \(\ce{CH_4}\)). Product side has 1 H per \(\ce{HCl}\). To balance H, we need 4 \(\ce{HCl}\) (since \(4 \times 1 = 4\) H). Thus, add a coefficient of 4 to \(\ce{HCl}\): \(\ce{CH_4 + Cl_2 \rightarrow CCl_4 + 4HCl}\) - **Chlorine (Cl):** Now, product side has \(4\) Cl (from \(\ce{CCl_4}\)) + \(4\) Cl (from \(4\ce{HCl}\)) = \(8\) Cl. Reactant side has 2 Cl per \(\ce{Cl_2}\). To get 8 Cl, we need \(4\ce{Cl_2}\) (since \(4 \times 2 = 8\) Cl). Thus, add a coefficient of 4 to \(\ce{Cl_2}\): \(\ce{CH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl}\) #### Step 3: Verify Balance Check all atoms: - **C:** 1 (reactant) = 1 (product) ✔️ - **H:** 4 (reactant, \(\ce{CH_4}\)) = 4 (product, \(4\ce{HCl}\)) ✔️ - **Cl:** 8 (reactant, \(4\ce{Cl_2}\)) = 8 (product, \(\ce{CCl_4}\) + \(4\ce{HCl}\)) ✔️ ### Relevant Knowledge Points #### 1. Law of Conservation of Mass In a chemical reaction, **mass is neither created nor destroyed**. This means the number of each type of atom must be equal on both the reactant and product sides (atoms are rearranged, not created or destroyed). #### 2. Balancing Chemical Equations To balance an equation: - **Count atoms** for each element on both sides. - Use **coefficients** (not subscripts) to adjust atom counts. Coefficients multiply all atoms in the formula they precede (e.g., \(3\ce{H_2O}\) has \(3 \times 2 = 6\) H atoms and \(3 \times 1 = 3\) O atoms). - Balance one atom at a time (start with atoms in the fewest compounds, e.g., C here). #### 3. Subscripts vs. Coefficients - **Subscripts** (e.g., the "4" in \(\ce{CH_4}\)) represent the number of atoms in a single molecule (e.g., \(\ce{CH_4}\) has 1 C and 4 H per molecule). - **Coefficients** (e.g., the "4" in \(4\ce{Cl_2}\)) represent the number of molecules (e.g., \(4\ce{Cl_2}\) means 4 molecules of \(\ce{Cl_2}\), with \(4 \times 2 = 8\) Cl atoms total). *Changing subscripts would change the substance (e.g., \(\ce{CH_2}\) is not methane), so only coefficients are adjusted.* ### Explanation of Knowledge Points (Professional + Easy to Understand) - **Law of Conservation of Mass:** Imagine baking a cake—you can’t create or destroy ingredients, only rearrange them. In chemistry, atoms work the same way: the total number of each atom must match on both sides of a reaction. - **Balancing Equations:** Think of it as a puzzle—you adjust the number of “molecule boxes” (coefficients) so that the total number of each atom (e.g., C, H, Cl) in the boxes matches on both sides. - **Subscripts vs. Coefficients:** Subscripts are like the “recipe” for one molecule (e.g., 1 C and 4 H in one \(\ce{CH_4}\) molecule). Coefficients are how many “copies” of that recipe you use (e.g., 4 copies of \(\ce{Cl_2}\) means 8 Cl atoms total). By following these steps and understanding the core concepts, you can balance any chemical equation!

Loading solution...