Question

Calculate the mass of carbon dioxide, CO2, produced from 0.85 mole of sucrose, C₆H₁₂O₆, assuming that oxygen, O₂ is in excess.

Answer 1

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

According to the reaction:

n(CO₂) = 6n(C₆H₁₂O₆) $= 6 \times 0.85 = 5.1 \;mol$

M(CO₂) = 44 g/mol

m(CO₂) $= 5.1 \times 44 = 224.4 \;g$

Answer: 224.4 g