Question

Photosynthesis allows plants to turn light energy into chemical energy by forming glucose from carbon dioxide and water: 6CO2(g) + 6H2O(l) → C6H12O6(g) + O2(g) ΔHrxn = 2802.8 kJ The glucose is then used to form complex carbohydrates that the plant needs to survive. A 1.70 lb sweet potato is approximately 73% water by mass. If the remaining mass is made up of carbohydrates derived from glucose (MW = 180.156 g/mol), how much carbon dioxide (MW = 44.01 g/mol) was needed to grow this sweet potato? g How much light energy does it take to grow the 1.70 lb. sweet potato if the efficiency of photosynthesis is 0.86%? kJ

Answer 1

305 g of CO₂

3.77 × 10⁵ kJ

Step-by-step explanation:

Let's consider the global reaction for photosynthesis.

6 CO₂(g) + 6 H₂O(l) → C₆H₁₂O₆(g) + 6 O₂(g) ΔHrxn = 2802.8 kJ

A 1.70 lb sweet potato is approximately 73% water by mass. If the remaining mass is made up of carbohydrates derived from glucose (MW = 180.156 g/mol), how much carbon dioxide (MW = 44.01 g/mol) was needed to grow this sweet potato?

Let's consider the following relations:

• The potato is 100%-73%=27% glucose by mass.
• 1 lb = 453.59 g.
• 6 moles of CO₂ produce 1 mole of glucose.
• The molar mass of glucose is 180.156 g/mol.
• The molar mass of carbon dioxide is 44.01 g/mol.

Then, for a 1.70 lb potato:

`1.70lbPotato.(27lbGlucose)/(100lbPotato) .(453.59gGlucose)/(1lbGlucose) .(1molGlucose)/(180.156gGlucose) .(6molCO_(2))/(1molGlucose) .(44.01gCO_(2))/(1molCO_(2)) =305gCO_(2)`

How much light energy does it take to grow the 1.70 lb. sweet potato if the efficiency of photosynthesis is 0.86%?

According to the enthalpy of the reaction, 2802.8 kJ are required to produce 1 mole of glucose. Then, for a 1.70 lb potato:

`1.70lbPotato.(27lbGlucose)/(100lbPotato) .(453.59gGlucose)/(1lbGlucose) .(1molGlucose)/(180.156gGlucose) .(2802.8kJ)/(1molGlucose) .(1)/(0.86\% ) =3.77* 10^(5) kJ`