Final answer:
To find the grams of CO₂ produced from the combustion of 2.45 mol of C₆H₆, multiply the moles of C₆H₆ by the stoichiometric ratio to get moles of CO₂ and then by the molar mass of CO₂ to obtain the mass in grams. The result is 647.33 grams of CO₂.
Step-by-step explanation:
To calculate the number of grams of CO₂ produced from the combustion of 2.45 mol of C₆H₆, use the stoichiometry of the balanced chemical equation provided.
The balanced chemical equation is C₆H₆(g) + 15O₂(g) → 6CO₂(g) + 3H₂O(g), which indicates that 1 mole of C₆H₆ produces 6 moles of CO₂.
First, calculate the moles of CO₂ produced by multiplying the moles of C₆H₆ by the molar ratio of CO₂ to C₆H₆ from the balanced equation:
2.45 mol C₆H₆ * (6 mol CO₂ / 1 mol C₆H₆) = 14.7 mol CO₂.
Next, use the molar mass of CO₂, which is 44.009 g/mol, to find the mass of 14.7 mol CO₂:
14.7 mol CO₂ * 44.009 g/mol CO₂ = 647.3323 g CO₂.
Therefore, 2.45 mol of C₆H₆ will produce 647.33 grams of CO₂ when combusted.