If 44.8L of carbon dioxide were produced, how many liters of oxygen must have reacted with glucose?
C6H12O6 + 3O2 ——-> 6CO2 + 6H2O
2 mols of CO2 for every mol of O2
since 22.4 liters per mol at STP for O2 and also for CO2
then 11.2 liters of O2
Whoops, sorry, not paying attention
44.8/2 = 22.4 liters CO2
To determine the number of liters of oxygen that reacted with glucose, we need to use the balanced equation for the combustion of glucose. The equation is as follows:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
From the equation, we can see that for every 6 moles of oxygen (O2) consumed, 6 moles of carbon dioxide (CO2) are produced. Therefore, the molar ratio of oxygen to carbon dioxide is 6:6, or simply 1:1.
Now, let's convert the given volume of carbon dioxide (44.8L) into moles. To do this, we need to know the density of carbon dioxide at the given conditions or the molar volume of carbon dioxide. Assuming standard temperature and pressure (STP), the molar volume of any ideal gas is approximately 22.4 liters per mole.
So, 44.8L of carbon dioxide is equal to:
44.8L / 22.4 L/mol = 2 moles of carbon dioxide (CO2)
Since the molar ratio of oxygen to carbon dioxide is 1:1, we can conclude that 2 moles of oxygen (O2) must have reacted with glucose.
To convert the moles of oxygen to liters, we will once again use the molar volume of an ideal gas at STP, which is 22.4 liters per mole.
Therefore, 2 moles of oxygen is equal to:
2 moles * 22.4 L/mol = 44.8 liters of oxygen (O2)
Hence, if 44.8 liters of carbon dioxide were produced, 44.8 liters of oxygen must have reacted with glucose.