# Problem Breakdown You're asked: **How much polyethylene is formed from 28 g of ethene, assuming 100% conversion via addition polymerization?** Let's solve step by step. --- ## 1. **Write the Polymerization Reaction** Ethene undergoes addition polymerization to form polyethylene: \[ n\,\text{C}_2\text{H}_4 \longrightarrow\, [\text{C}_2\text{H}_4]_n \] Here, \( n \) molecules of ethene (C₂H₄) combine to form a polymer chain of polyethylene. --- ## 2. **Molar Mass Calculations** - **Ethene (C₂H₄):** \( 2 \times 12.01\,(\text{C}) + 4 \times 1.008\,(\text{H}) = 24.02 + 4.032 = 28.052\,\text{g/mol} \) (We'll use 28 g/mol for simplicity.) - **Polyethylene repeat unit:** Each repeat unit \((\text{C}_2\text{H}_4)\) has mass 28 g/mol. --- ## 3. **Mole Calculation** Given mass of ethene = 28 g Number of moles: \[ \text{Moles of ethene} = \frac{28\,\text{g}}{28\,\text{g/mol}} = 1\,\text{mol} \] --- ## 4. **Polymerization Stoichiometry** Each \( n \) moles of ethene gives 1 mole of polyethylene of degree of polymerization \( n \). But total mass is conserved (no atoms lost in addition polymerization). - **Total mass of monomer = Total mass of polymer** (for 100% conversion). --- ## 5. **Final Calculation** \[ \text{Mass of polyethylene} = \text{mass of ethene used} = 28\,\text{g} \] --- ## 6. **Answer** > **The mass of polyethylene obtained from 28 g of ethene (100% conversion) is:** > > \[ > \boxed{28\,\text{g}} > \] --- ### **Summary Table** | Given | Value | |------------------|---------------| | Mass of ethene | 28 g | | Molar mass C₂H₄ | 28 g/mol | | Moles of C₂H₄ | 1 mol | | Mass of polymer | 28 g | --- **Key Point:** In addition polymerization (no byproducts), the mass of polymer equals the mass of monomer used.