how many moles of glucose can be burned "biologically" when 10.0 mol of oxygen is available? C6 H12 O6 + 6CO2 -> 6CO2 +6H2O
I think your equation is really, really wrong. Try
C6 H12 O6 + 6O2 -> 6CO2 +6H2O
10 mol O2? then 1/6 of 10 moles is moles of glucose
To determine the moles of glucose that can be burned biologically using 10.0 mol of oxygen, we need to apply the stoichiometry of the chemical equation provided.
The balanced equation for the combustion of glucose is:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
From the equation, we can see that for every mole of glucose (C6H12O6), 6 moles of oxygen (O2) are required. Therefore, the ratio of moles of oxygen to moles of glucose is 6:1.
So, if we have 10.0 mol of oxygen available, we can calculate the moles of glucose burned using the following ratio:
10.0 mol O2 * (1 mol glucose / 6 mol O2) = 1.67 mol glucose
Thus, approximately 1.67 moles of glucose can be burned biologically when 10.0 mol of oxygen is available.