Hexane \( \left(\mathrm{C}_{6} \mathrm{H}_{14}\right) \) burns with oxygen as part of a combustion reaction. The products are carbon dioxide and water. How many grams of hexane are required to produce \( 3.98 \times 10^{-3} \mathrm{~g} \) of wate Antifreeze is made up of ethylene glycol \( \left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{3}\right) \) can be produced from a synthesis reaction

**Step 1. Write the balanced combustion equation for hexane.** Hexane \(\mathrm{C}_{6}\mathrm{H}_{14}\) burns in oxygen to produce carbon dioxide and water. A properly balanced equation is \[ 2\,\mathrm{C}_{6}\mathrm{H}_{14} + 19\,\mathrm{O}_2 \longrightarrow 12\,\mathrm{CO}_2 + 14\,\mathrm{H}_2\mathrm{O} \] This equation shows that \(2\) moles of hexane produce \(14\) moles of water. Therefore, \(1\) mole of hexane produces \(7\) moles of water. **Step 2. Calculate the number of moles of water produced.** The mass of water given is \[ 3.98 \times 10^{-3} \text{ g} \] The molar mass of water, \(\mathrm{H}_2\mathrm{O}\), is approximately \[ 18.02\, \mathrm{g/mol} \] Thus, the number of moles of water is \[ n(\mathrm{H}_2\mathrm{O}) = \frac{3.98 \times 10^{-3}\,\mathrm{g}}{18.02\,\mathrm{g/mol}} \approx 2.21 \times 10^{-4}\,\mathrm{mol} \] **Step 3. Determine the number of moles of hexane required.** Since \(1\) mole of hexane produces \(7\) moles of water, the moles of hexane required are \[ n(\text{hexane}) = \frac{n(\mathrm{H}_2\mathrm{O})}{7} = \frac{2.21 \times 10^{-4}\,\mathrm{mol}}{7} \approx 3.16 \times 10^{-5}\,\mathrm{mol} \] **Step 4. Calculate the mass of hexane required.** The molar mass of hexane \(\mathrm{C}_{6}\mathrm{H}_{14}\) is calculated as follows: \[ \text{Molar mass} = 6 \times 12.01 + 14 \times 1.008 \approx 72.06 + 14.112 \approx 86.172\,\mathrm{g/mol} \] Thus, the mass of hexane needed is \[ \text{Mass} = n(\text{hexane}) \times \text{Molar mass} \approx 3.16 \times 10^{-5}\,\mathrm{mol} \times 86.17\,\mathrm{g/mol} \approx 2.73 \times 10^{-3}\,\mathrm{g} \] **Answer for the combustion of hexane:** Approximately \(2.73 \times 10^{-3}\) grams of hexane are required to produce \(3.98 \times 10^{-3}\) grams of water. **Note on the Antifreeze Problem:** The problem statement regarding antifreeze production from ethylene glycol \(\mathrm{C}_2\mathrm{H}_6\mathrm{O}_3\) mentions a synthesis reaction but does not provide sufficient details (e.g., reaction conditions, amounts of reactants or products) to carry out a specific calculation. Additional information would be needed to solve that part.