1. Write the balanced chemical equation:

C₆H₆ (l) + 15/2 O₂ (g) → 3 CO₂ (g) + 3 H₂O (g)

2. Calculate the theoretical yield of CO₂:

* Molar mass of benzene (C₆H₆): 78.11 g/mol

* Molar mass of CO₂: 44.01 g/mol

* Stoichiometry: From the balanced equation, 1 mole of benzene produces 3 moles of CO₂.

* Grams of benzene to moles: 7.80 g C₆H₆ / 78.11 g/mol = 0.1 mol C₆H₆

* Moles of benzene to moles of CO₂: 0.1 mol C₆H₆ * (3 mol CO₂ / 1 mol C₆H₆) = 0.3 mol CO₂

* Moles of CO₂ to grams of CO₂: 0.3 mol CO₂ * 44.01 g/mol = 13.20 g CO₂

3. Calculate the percent yield:

* Percent yield = (actual yield / theoretical yield) * 100%

You haven't provided the actual yield of CO₂ in the problem. To calculate the percent yield, you'll need the actual amount of CO₂ produced in the experiment.

Let's assume you obtained 10.5 grams of CO₂ in your experiment:

* Percent yield = (10.5 g CO₂ / 13.20 g CO₂) * 100% = 79.5%

Therefore, if the actual yield of CO₂ is 10.5 grams, the percent yield is 79.5%.