Question

If 9.25 liters of benzene (C₆H₆) at standard temperature and pressure (STP) is burned in air, what is the mass of carbon dioxide formed? Round your answer to two decimals.

Answer 1

The mass of carbon dioxide formed when 9.25 liters of benzene is burned at STP is 108.92 g, calculated using stoichiometry and the molar volume at STP.

Step-by-step explanation:

To calculate the mass of carbon dioxide (CO₂) formed from burning benzene (C₆H₆) at STP, we use stoichiometry based on the balanced chemical equation for the combustion of benzene:

2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)

First, we convert the given volume of benzene to moles using the molar volume at STP (22.414 L mol⁻¹):

Moles of C₆H₆ = 9.25 L / 22.414 L mol⁻¹ = 0.4125 moles

From the stoichiometry of the balanced equation, the mole ratio of benzene to CO₂ is 2:12, or 1:6. Using this ratio, we calculate the moles of CO₂ produced:

Moles of CO₂ = 0.4125 moles * 6 = 2.475 moles

Now, we use the molar mass of CO₂ (44.01 g mol⁻¹) to find the mass:

Mass of CO₂ = 2.475 moles * 44.01 g mol⁻¹ = 108.92 g

Therefore, the mass of CO₂ formed when 9.25 liters of benzene is burned at STP is 108.92 g, rounded to two decimals.