1. Write the balanced chemical equation:

The combustion of 1-heptanol (C7H16O) produces carbon dioxide (CO2) and water (H2O):

C7H16O + 10O2 → 7CO2 + 8H2O

2. Calculate the molar mass of 1-heptanol and carbon dioxide:

* Molar mass of 1-heptanol (C7H16O) = (7 * 12.01 g/mol) + (16 * 1.01 g/mol) + (1 * 16.00 g/mol) = 116.21 g/mol

* Molar mass of carbon dioxide (CO2) = (1 * 12.01 g/mol) + (2 * 16.00 g/mol) = 44.01 g/mol

3. Determine the mole ratio between 1-heptanol and carbon dioxide:

From the balanced equation, 1 mole of 1-heptanol produces 7 moles of carbon dioxide.

4. Calculate the moles of 1-heptanol:

* Moles of 1-heptanol = (mass of 1-heptanol) / (molar mass of 1-heptanol)

* Moles of 1-heptanol = (23.5 g) / (116.21 g/mol) = 0.202 mol

5. Calculate the moles of carbon dioxide produced:

* Moles of CO2 = (moles of 1-heptanol) * (mole ratio of CO2 to 1-heptanol)

* Moles of CO2 = (0.202 mol) * (7 mol CO2 / 1 mol 1-heptanol) = 1.414 mol

6. Calculate the mass of carbon dioxide produced:

* Mass of CO2 = (moles of CO2) * (molar mass of CO2)

* Mass of CO2 = (1.414 mol) * (44.01 g/mol) = 62.2 g

Therefore, 62.2 grams of carbon dioxide are produced when 23.5 grams of 1-heptanol react with excess oxygen.