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