17. Our life on earth would not exist without photosynthesis-the conversion of c

Question

17. Our life on earth would not exist without photosynthesis-the conversion of carbon dioxide into glucose (CoH120is) by capturing the energy of the sun, plus oxygen (as a side product) essential for animals. 14 pts. a) Using the information in the table calculate the energy absorbed from sunlight (AHo) for this process. Compound AH CsH120s6)1274.5 kJ/mol co0) 285,8 H2O 393.5 kJ/mol -285.8 kJ/mol Energy from the sun needed for this process b) How many grams of glucose could be produced from 100.0 grams of carbon dioxide? Mass of glucose produced from 100.0 grams of CO2 = c) If a plant had available for photosynthesis 100.0 g of CO2 and 100.0 g H20 i) What is the limiting reagent? Limiting reagent i) What is the maximum amount of glucose that this plant can produce? Maximum amount of glucose-

Explanation / Answer

Ans. #a. Total dH0f of reactant = 6 mol x dH0f CO2(g) + 6 mol x dH0f H2O(l)

                                    = 6 mol x (-393.5 kJ/mol) + 6 mol x (-285.8 kJ/mol)

                                    = - 4075.8 kJ

# Total dG0f of products = 6 mol x dF0f O2(g) + 1 mol x dH0f C6H12O6(g)

                                    = 6 mol x (0.0 kJ/mol) + 1 mol x (-1274.5kJ/mol)

                                    = -1274.5 kJ

# Now, dH0Rxn = (Total dH0f of products) – (Total dH0f of reactant)

                                    = -1274.5 kJ – (-4075.8 kJ)

                                    = +2801.3 kJ

# Therefore, energy absorbed during the reaction, dH0Rxn = 2801.3 kJ

#b. Moles of CO2 = Mass / MW = 100.0 g / (44.0 g/ mol) = 2.273 mol

Following stoichiometry, 6 mol CO2 forms 1 mol glucose.

So,

            Moles of glucose formed = (1/6) x Moles of CO2 = (1/6) x 2.273 mol

= 0.3788 mol

Now,

Mass of glucose formed = Moles x MW = 0.3788 mol x (180.15 g/ mol)

                                                = 68.25 g

#c. Step 1: Determine limiting reactant:

Moles of CO2 = 2.273 mol

Moles of H2O = 100.0 g / (18.0 g/ mol) = 5.56 mol

# Following stoichiometry of balanced reaction, 6 mol CO2 reacts with 6 mol H2O, i.e. 1 mol CO2 reacts with 1 mol H2O.

Since given moles of H2O is greater than that of CO2, water is the reagent in excess.

Hence, CO2 is the limiting reactant.

# Step 2: Determine yield:

Formation of product follows the stoichiometry of limiting reactant.

Following stoichiometry, 6 mol CO2 forms 1 mol glucose.

So,

            Moles of glucose formed = (1/6) x Moles of CO2 = (1/6) x 2.273 mol

= 0.3788 mol

Now,

Mass of glucose formed = Moles x MW = 0.3788 mol x (180.15 g/ mol)

                                                = 68.25 g

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