If a plant has 88.0 g carbon dioxide and 64.0g water available for photosynthesis determine:

Question

a the limiting reactant
b. the excess reactant and the mass in excess
c. the mass of glucose produced

Answer 1

6CO2 + 6H2O ==> C6H12O6 + 6O2
mols CO2 = grams/molar mass
mols H2O = grams/molar mass
You need to redo the math because these are only approx but I obtained approx.
About 2 mols CO2; about 3.5

Convert 2 mols CO2 to mols glucose; that's 2 x (1 mol glucose/6 mol CO2) = about 0.33 mol glucose
Convert 3.5 mol H2O to mols glucose; that's about 3.5 x (1 mol glucose/5 mols H2O) = about 0.6.
You see the value is not the same for mols glucose which means one of the numbers is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. So CO2 is the limiting reagent. grams glucose = mols glucose x molar mass glucose = approx 0.33 x molar mass glucose = ?

For part b, the excess reagent obviously is H2O. How much. Use the same stoichiometry to determine how much H2O was used. That's approx 0.33 mols CO2 x (6 mols CO2/6 mol H2O) = about 0.33 mols H2O used.
You had approx 0.6 mol initially, you used about 0.33 so the difference is the amount H2O left in mols. To find grams, g = mols x molar mass.