How many moles of glucose , C6H12O6, can be "burned" biologically when 10.0 mol of oxygen is available.
C6H12O6(s) + 602(g)---6CO2(g) + 6H2O(l)
1.67 mol
Solving for glucose. 10.0 mol divided by 6 O2 is the answer (1.67)
To determine the number of moles of glucose that can be burned biologically, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that the ratio of C6H12O6 to O2 is 1:6. This means that for every 1 mole of glucose reacted, we need 6 moles of oxygen.
Therefore, if we have 10.0 moles of oxygen available, we can burn (10.0 mol) / (6 mol O2 per 1 mol C6H12O6) = 1.67 moles of glucose.
Thus, 1.67 moles of glucose can be burned when 10.0 moles of oxygen is available.
To determine the number of moles of glucose that can be burned biologically, we need to first find the balanced chemical equation for the reaction. The equation you have provided is already balanced:
C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(l)
From the balanced equation, we can see that 1 mol of glucose reacts with 6 mol of oxygen to produce 6 mol of carbon dioxide and 6 mol of water.
Given that 10.0 mol of oxygen is available, we can use the stoichiometric ratios from the balanced equation to determine the number of moles of glucose that can be burned.
Since there is a 6:6 ratio between oxygen and glucose, we can conclude that for every 6 mol of oxygen, 1 mol of glucose is burned.
So, if 6 mol of oxygen can burn 1 mol of glucose, then 10.0 mol of oxygen can burn:
10.0 mol O2 * 1 mol glucose/6 mol O2 = 1.67 mol of glucose
Therefore, 10.0 mol of oxygen can burn approximately 1.67 mol of glucose.