**How many grams of **CO_{2}** are produced when 2 moles of C**_{3}H_{8} are burned in excess oxygen?

**Solution:** The balanced chemical equation for the combustion of propane (C_{3}H_{8}) is:

C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O

According to the equation, 1 mole of C_{3}H_{8} produces 3 moles of CO_{2}. Therefore, 2 moles of C_{3}H_{8} will produce:

2 moles C_{3}H_{8} × (3 moles CO_{2} / 1 mole C_{3}H_{8}) = 6 moles CO_{2}

To convert moles of CO_{2} to grams, we need to use the molar mass of CO_{2}, which is 44 g/mol:

Mass of CO_{2} = 6 moles × 44 g/mol = 264 g

Therefore, 264 grams of CO_{2} are produced when 2 moles of C_{3}H_{8} are burned in excess oxygen.

I hope these problems help you practice for your NEET exams!