Asked by Jordayna
If possibly please show work
If 1.00 kg of glucose (5.55 mol) is reacted with 34.0 mol oxygen during the process of cellular respiration, which reactant is limiting? Why does your answer seem logical?
If 1.00 kg of glucose (5.55 mol) is reacted with 34.0 mol oxygen during the process of cellular respiration, which reactant is limiting? Why does your answer seem logical?
Answers
Answered by
oobleck
what is the reaction equation?
If more than 34.0/5.55 = 6.18 moles of O2 are used for each mole of glucose, then the O2 will run out first.
If more than 34.0/5.55 = 6.18 moles of O2 are used for each mole of glucose, then the O2 will run out first.
Answered by
DrBob222
C6H12O6 + 6O2 → 6CO2 + 6H2O + Energy
mols C6H12O6 = 5.55 mols
5.55 mols C6H12O6 will require 6(5.55 = 33.3 mols O2.and you have that much 9at 34) so the reaction will occur as written and glucose will be the limiting reagent.
mols C6H12O6 = 5.55 mols
5.55 mols C6H12O6 will require 6(5.55 = 33.3 mols O2.and you have that much 9at 34) so the reaction will occur as written and glucose will be the limiting reagent.
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