In the combustion analysis of 0.1127g of glucose (c6H12O6), what mass, in grams, of CO2 would be produced?
by 
angeybutt
ANSWERS: 2
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Well, let's see. The molar mass of glucose is around 180 gm/mol and the molar mass of CO2 is close to 44 gm/mol. You should already know how these molar masses were calculated. We also obtain a balanced combustion reaction: C6H12O6+ 6O2--->6CO2+ 6H2O. This tells us for every mol of glucose consumed, there are 6 mols of CO2 produced. Divide the molar mass of glucose into 0.1127 gm of glucose to obtain mols of glucose. If you multiply the mols of glucose by the stoichiometric ratio of CO2 produced (6), it yields the mols of CO2 produced. To find the mass of CO2 produced, divide the mols of CO2 (calculated from above) by the molar mass of CO2 and voila! I can't make it any easier without giving you the answer. Good luck!
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Glucose and oxygen will react to produce CO2 and H2O. In glucose, the hydrogen/oxygen ratio is 2/1 so the water accounts for all of the Hydrogen. That leaves carbon to react with external oxygen at a ratio of 6 to 1 (glucose has 6 carbon, 6 02 molecules needed to react with these). So figure out the mass of a mole of glucose (I got 180.18g) and divide by the given 0.1127g to get the number of moles of glucose. Since it's 6 to 1 oxygen, take the number of moles of glucose times 6 to get the moles of CO2. Multiply that number times the mass of a mole of CO2 (I used 44.01g per mole). The result is 0.1652g of CO2.
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