The combustion of glucose (C6H12O6) with oxygen gas produces carbon dioxide and water. This process releases 2803 kJ per mole of glucose. When 3.00...

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 +...

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.