To answer this question, we must first identify how much oxygen is reacted with the glucose to make it burn. To start, I will write the chemical equation, but with variables for the unknown quantities.

C6H12O6 + xO2 => yCO2 + zH2O

Here, x, y, and z are the variables....

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To answer this question, we must first identify how much oxygen is reacted with the glucose to make it burn. To start, I will write the chemical equation, but with variables for the unknown quantities.

C6H12O6 + xO2 => yCO2 + zH2O

Here, x, y, and z are the variables. I will start balancing with the carbon. Because there are 6 carbon in the glucose, there must be 6 carbons in the CO2.

C6H12O6 + xO2 => 6CO2 + zH2O

Now, I balance the hydrogen. Again, because there are twelve hydrogen in the glucose, there must be twelve on the opposite side. Notice that there are 6 H2O rather than 12.

C6H12O6 + xO2 => 6CO2 + 6H2O

Now we just add up the oxygen on the right, and subtract the oxygen in the glucose. I count 18 on the right, and 6 in the glucose, so there must be 12 oxygen atoms, or 6 oxygen molecules.

C6H12O6 + 6O2 => 6CO2 + 6H2O

Now, we can say that it takes six moles of oxygen to burn one mole of glucose. All that is left is a simple fraction and a unit conversion.

`(6 mol O2)/(2803 kj)=(3 mol O2)/x`

Solving for x gives 1401.5 kj, and converting to kcal gives 334.74 kcal.