Question

15. In photosynthesis, plants convert carbon dioxide and water into oxygen and glucose, C6H1206. If 75.2 g of glucose are obtained when 131 g of CO2 react in the presence of excess H2O, what is the percent yield of the reaction?

Answer 1

6CO_2 (g) + 6H_2O (l) rightarrow 6O_2 (g) + C_6 H_12 O_6 (g) Mass of CO_2 = 131 g Mass of C_6 H_12 O_6 obtained = 75.2q so, Actual yield = 75.2 g. Molar mass of CO_2 = 12 + 2 times 16 = 12 + 32 = 44 g/mol. Molar mass of C_6 H_12 O_12 O_6 = 6 times 12 + 12 times 1 + 6 times 16 = 72 + 12 + 96 = 180 g/mol. So, 6 times 44 g CO_2 = 180 g C_6 H_12 O_6 131 g CO_2 = 180/6 times 44 times 131 g C_6 H_12 O_6 = 89.3 g C_6 H_12 O_6 Theoretical yield = 89.3 g % yield = Actual yield/Theoretical yield times 10 = 75.2/89.3 times 100 = 84.2%

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