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Next. Given the three equations below, what is the heat of reaction for the production of glucose, C6H12O6, as described by this equation 6C(s) + 6H2(g) + 3O2(g) â C6H12O6(s) C(s) + O2(g) â CO2(g), âH = â393.51 kJ H2(g) + Â½ O2(g) â H2O(l), âH = â285.83 kJ C6H12O6(s) + 6O2(g) â 6CO2(g) + H2O(l), âH = â2803.02...ext. Given the three equations below, what is the heat of reaction for the production of glucose, C6H12O6, as described by this equation 6C(s) + 6H2(g) + 3O2(g) â C6H12O6(s) C(s) + O2(g) â CO2(g), âH = â393.51 kJ H2(g) + Â½ O2(g) â H2O(l), âH = â285.83 kJ C6H12O6(s) + 6O2(g) â 6CO2(g) + H2O(l), âH = â2803.02...
One molecule of glucose (C6H12O6) is equal to one mole of glucose. To get your answer, you need to turn grams into moles. To get the number of grams for glucose, go to the periodic table to find out the weights for carbon, hydrogen and oxygen.Carbon is 12, hydrogen is 1 and oxygen is 16.he total is 180 grams/mole, which is the weight of one molecule (mole) of glucose. To find out the number of moles, divide 300 grams by 180 grams/mole. The grams cancel each other out and you are left with 1.67 moles. Pandra Â· 7 years ago. Thumbs up. 0. Thumbs down. 0. 1 comment. Loading ...