How many moles of glucose, C6H12O6, can be "burned" biologically when 10.0 mol of oxygen is available?
1.67 mol
in a balanced equation there 6 mol of oxygen for every 1 mol of glucose; so ratio is 6 to 1.
10 mol oxygen/ 6 = 1.67...
See example below.
60.0
To determine the number of moles of glucose that can be burned biologically when 10.0 mol of oxygen is available, we need to consider the balanced chemical equation for the combustion of glucose:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
From the balanced equation, we can see that 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of carbon dioxide and 6 moles of water.
Since we know that 10.0 mol of oxygen is available, we can use the stoichiometry of the equation to determine how many moles of glucose would be required for complete combustion.
Using a ratio, we can set up a proportion:
6 moles of oxygen / 1 mole of glucose = 10.0 moles of oxygen / x moles of glucose
By cross-multiplying and solving for x, we can find the number of moles of glucose:
x = 10.0 moles of oxygen * (1 mole of glucose / 6 moles of oxygen)
x = 10.0 / 6
x = 1.67 moles
Therefore, approximately 1.67 moles of glucose can be burned biologically when 10.0 mol of oxygen is available.