A reactor is supplied with C6H12O6 and the following reactions occur C6H12O6 -> 2C2H5OH + 2CO2 and C6H12O6 –> 2C2H3CO2H + 2H2O, 360 grams of pure glucose is supplied and 44 grams of CO2 was produced, what 90 grams of glucose left unreacted, how many grams of water was produced?

Correct Answer: 9
Weight of glucose reacted in first reaction = ½*(66/44)*180 = 135 grams => Weight of glucose reacted in second reaction = 360 – 180 – 135 = 45 grams, => moles of water produced = 2*(45/180)*18 = 9 grams.