Photosynthesis
Afra Khanani
Honors Chemistry
Period 6
March 31st
PROBLEM:
A solution containing 6720 mg of H20 is added to
another containing 10.67 Liters of CO2 at STP.
Determine which reactant was in excess, as well as
the number of grams over the amount required by
the limiting species. Also, find the number of
molecules of glucose that precipitated as well, as
the theoretical and percent yield of glucose if 10.22
g C6H12O6 was obtained.
STEP 1
Write and balance the equation
STEP 1
Write and balance the equation
__ H20 + __ CO2
REACTANTS
__ C6H12O6 + __ 02
PRODUCTS
STEP 1
Write and balance the equation
__ H20 + __ CO2
REACTANTS
__ C6H12O6 + __ 02
PRODUCTS
Water + Carbon Dioxide (+ energy) = Glucose + Oxygen
STEP 2
Start with one of the knowns (convert mg to g)
6720 mg of H20 (Given)
STEP 2
Start with one of the knowns (convert mg to g)
6720 mg of H20 (Given)
6720 mg H20
1 gram H20
1000 milligrams H20
STEP 3
Convert grams to moles
STEP 3
Convert grams to moles
6.72 g H20
1 mole H20
18 grams H20
STEP 4
Convert mole to moles
STEP 4
Convert mole to moles
6 H20 + 6 CO2
C6H12O6 + 6 02
STEP 4
Convert mole to moles
6 H20 + 6 CO2
.373 mole H20
C6H12O6 + 6 02
1 mole C6H12O6
6 mole H20
STEP 5
Convert moles to grams
STEP 5
Convert moles to grams
.062 mole C6H1206
180 grams C6H1206
1 mole C6H1206
You have figured out that:
6720 mg of H20 = 11.16 grams C6H12O6
Now figure out:
10.67 L CO2 = ? grams C6H12O6
STEP 1
Convert L at STP to moles
10.67 L of CO2 (Given)
STEP 1
Convert L at STP to moles
10.67 L of CO2 (Given)
10.67 L CO2
1 mole CO2
22.4 Liters CO2
STEP 2
Convert moles to moles
6 H20 + 6 CO2
C6H12O6 + 6 02
STEP 2
Convert moles to moles
6 H20 + 6 CO2
.476 mole CO2
C6H12O6 + 6 02
1 mole C6H12O6
6 mole CO2
STEP 3
Convert moles to grams
STEP 3
Convert moles to grams
.079 mole C6H12O6
180 grams C6H12O6
1 mole C6H12O6
You have figured out that:
6720 mg of H20 = 11.16 grams C6H12O6
10.67 L of CO2 = 14.22 mol C6H12O6
CO2 is the excess reactant
You have figured out that:
6720 mg of H20 = 11.16 grams C6H12O6
10.67 L of CO2 = 14.22 mol C6H12O6
CO2 is the excess reactant
How much excess CO2 ?
(In grams)
You have figured out that:
6720 mg of H20 = 11.16 grams C6H12O6
10.67 L of CO2 = 14.22 mol C6H12O6
CO2 is the excess reactant
How much excess CO2 ?
(In grams)
14.22 grams – 11.16 grams = 3.06 grams CO2 in excess
What’s Next?
Find the number of molecules of glucose
that precipitated.
STEP 1
Convert moles to molecules
0.62 mole of C6H12O6 (Found)
STEP 1
Convert moles to molecules
0.62 mole of C6H12O6 (Found)
0.62 moles C6H12O6
6.02 x 1023 C6H12O6
1 mole C6H12O6
RESTATING THE QUESTION:
Find the number of molecules of glucose
that precipitated.
RESTATING THE QUESTION:
Find the number of molecules of glucose
that precipitated.
THEORETICAL & PERCENT YIELD
Find theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
THEORETICAL & PERCENT YIELD
Find theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
Theoretical Yield: (Already Found) 11.16 grams
THEORETICAL & PERCENT YIELD
Find theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
Theoretical Yield: (Already Found) 11.16 grams
Percent Yield: ACTUAL YIELD/THEORETICAL x 100
THEORETICAL & PERCENT YIELD
Find theoretical percent yield of C6H12O6
(Actual amount of Glucose obtained was 10.22 as stated before)
Theoretical Yield: (Already Found) 11.16 grams
Percent Yield: ACTUAL YIELD/THEORETICAL x 100
PERCENT YIELD: 10.22/11.16 x 100 = 92%