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

Answer:

305 g of CO₂

3.77 × 10⁵ kJ

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:

$1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{6molCO_{2}}{1molGlucose} .\frac{44.01gCO_{2}}{1molCO_{2}} =305gCO_{2}$

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:

$1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{2802.8kJ}{1molGlucose} .\frac{1}{0.86\% } =3.77\times 10^{5} kJ$