a. If 5.57 g of glucose are produced through photosynthesis, what is the minimum number of moles of carbon dioxide and of water that were used?
b. If the 5.57 g of glucose is an 80.2% yield, and assuming that carbon dioxide was the limiting reagent, how many moles of carbon dioxide were available for the reaction?
To answer these questions, we need to understand the chemical equation for the process of photosynthesis and some basic concepts of stoichiometry.
a. In the process of photosynthesis, glucose is produced from carbon dioxide (CO2) and water (H2O) using light energy. The chemical equation for photosynthesis is:
6CO2 + 6H2O -> C6H12O6 + 6O2
From this balanced equation, we can see that 1 mole of glucose (C6H12O6) is produced from 6 moles of carbon dioxide and 6 moles of water.
To find the minimum number of moles of carbon dioxide and water used, we can use the molar mass of glucose to convert the given mass (5.57 g) into moles.
The molar mass of glucose (C6H12O6) is:
(6*C) + (12*H) + (6*O) = 72 g/mol (carbon: 12 g/mol, hydrogen: 1 g/mol, oxygen: 16 g/mol)
Number of moles of glucose = mass / molar mass
Number of moles of glucose = 5.57 g / 72 g/mol
Number of moles of glucose = 0.0772 mol
From the balanced equation, we know that 1 mole of glucose requires 6 moles of carbon dioxide and 6 moles of water.
Therefore, the minimum number of moles of carbon dioxide and water used would be:
Number of moles of carbon dioxide = 0.0772 mol * 6 = 0.4632 mol
Number of moles of water = 0.0772 mol * 6 = 0.4632 mol
So, the minimum number of moles of carbon dioxide and water used are both 0.4632 mol.
b. Assuming that carbon dioxide is the limiting reagent, we can determine the amount of carbon dioxide consumed in the reaction using the given mass of glucose (5.57 g) and its yield (80.2%).
Since the yield is given as a percentage, it represents the fraction of the theoretical yield obtained. Hence, to find the theoretical yield of glucose, we can use the given mass and the yield percentage:
Theoretical yield of glucose = mass of glucose / yield percentage
Theoretical yield of glucose = 5.57 g / 80.2%
Theoretical yield of glucose = 5.57 g / 0.802
Theoretical yield of glucose = 6.933 g
From the balanced equation, we know that 1 mole of glucose (180 g/mol) gives 6 moles of carbon dioxide (6 * 44 g/mol).
Number of moles of glucose = mass / molar mass
Number of moles of glucose = 6.933 g / 180 g/mol
Number of moles of glucose = 0.0385 mol
Since the stoichiometry between glucose and carbon dioxide is 1:6, the moles of carbon dioxide would be:
Number of moles of carbon dioxide = number of moles of glucose * 6
Number of moles of carbon dioxide = 0.0385 mol * 6
Number of moles of carbon dioxide = 0.231 mol
Therefore, if the 5.57 g of glucose is an 80.2% yield and carbon dioxide is the limiting reagent, then there were approximately 0.231 moles of carbon dioxide available for the reaction.