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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...
You know the formula of glucose C6H12O6 and molecular weight 180, so 18 g will be 18/180 = 0.1 mole Now Avogadro No. says one mole has 6.022X10^23 particles, so 0.1 mole glucose will be 0.1*6.022*10^23 molecules.Each molecule contains 6 carbon atoms so no. of carbon atoms will be 6*0.1*6.022*10^23= 3.6132*10^23.ou know the formula of glucose C6H12O6 and molecular weight 180, so 18 g will be 18/180 = 0.1 mole Now Avogadro No. says one mole has 6.022X10^23 particles, so 0.1 mole glucose will be 0.1*6.022*10^23 molecules. Each molecule contains 6 carbon atoms so no. of carbon atoms will be 6*0.1*6.022*10^23= 3.6132*10^23.