<span>so your ratio for the equation is 1 glucose + 6 oxygen = 6 carbon dioxide + 6 water
those ratios will all be in moles, and will all be based on the number of moles of your limiting reagent. We assume that you have unlimited oxygen, so glucose must be limiting. The molar mass for glucose is 180 grams (you can get more accurate if you need, this is just the number I am using).
100 g / 180 g/mol = 0.556 moles of glucose. since we will need six times the moles of glucose in oxygen... 6 x .556 = 3.33 moles of O2 multiply by the molar mass of O2 for the mass of oxygen needed... 3.33 x 32 = 106.67 grams of O2 since you are forming six times the moles of glucose in CO2 and H2O, you are forming 3.33 moles of each of these compounds as well.</span>
