Summary of problem
Photosynthesis uses CO2 & H20 to make C6H12O2. If a plant has 99 grams of CO2 & 68 grams of H2O for photosynthesis, what's the balanced equation, the mass of glucose made, & the mass of excess?
1) balance the equation by making sure you have the same number of carbon, oxygen, and hydrogen atoms on both the reactants and products side.
2) find the limiting reagent the reactant (CO2 or H2O) that produce the least amount of glucose
3) find the difference in the grams of glucose produced from the reactants
http://www.iun.edu/~cpanhd/C101webnotes/quantchem/limitingreagent.html
I can't get EXACTLY 68 grams of water; dunno how to change equation without using a fraction.
Add O2 on the product side and now balance the equation.
CO2 + H20 -> C6H12O2 + O2
To answer this question, we need to follow a step-by-step approach:
Step 1: Write the balanced equation for photosynthesis.
Step 2: Calculate the molar mass of CO2 and H2O.
Step 3: Determine the limiting reactant.
Step 4: Calculate the theoretical mass of glucose produced.
Step 5: Calculate the mass of the excess reactant.
Let's go through each step in detail:
Step 1: Write the balanced equation for photosynthesis.
The balanced equation for photosynthesis can be represented as:
6CO2 + 6H2O -> C6H12O6 + 6O2
Step 2: Calculate the molar mass of CO2 and H2O.
The molar mass of CO2 (carbon dioxide) is approximately 44 grams/mol (12 grams/mol for carbon and 32 grams/mol for oxygen).
The molar mass of H2O (water) is approximately 18 grams/mol (2 grams/mol for hydrogen and 16 grams/mol for oxygen).
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we need to compare the number of moles of each reactant with the stoichiometric coefficients in the balanced equation.
First, convert the given mass of CO2 and H2O to moles:
Number of moles of CO2 = 99 g / 44 g/mol ≈ 2.25 mol
Number of moles of H2O = 68 g / 18 g/mol ≈ 3.78 mol
Now, compare the moles of CO2 and H2O with the stoichiometric coefficients in the balanced equation:
CO2: 6 moles required for the reaction
H2O: 6 moles required for the reaction
Since both CO2 and H2O have more than enough moles to fully react, neither is limiting. Therefore, there is no limiting reactant in this case.
Step 4: Calculate the theoretical mass of glucose produced.
Using the balanced equation, we see that for every 6 moles of CO2 and 6 moles of H2O used, 1 mole of C6H12O6 (glucose) is produced.
The molar mass of glucose is approximately 180 grams/mol.
Since there is no limiting reactant, we can use either CO2 or H2O to calculate the theoretical mass of glucose. Let's use CO2:
Number of moles of CO2 = 2.25 mol (as calculated earlier)
Number of moles of glucose produced = 2.25 mol / 6 = 0.375 mol
Mass of glucose produced = Number of moles of glucose produced × Molar mass of glucose
Mass of glucose produced = 0.375 mol × 180 g/mol ≈ 67.5 grams
Step 5: Calculate the mass of the excess reactant.
Since there is no limiting reactant, there is no excess reactant in this case.
In summary:
- The balanced equation for photosynthesis is 6CO2 + 6H2O -> C6H12O6 + 6O2
- The mass of glucose made is approximately 67.5 grams.
- There is no mass of excess reactant in this case.