STOICHIOMETRY:

Afra KhananiHonors ChemistryPeriod 6March 31st

Photosynthesis

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 1Write and balance the equation

STEP 1Write and balance the equation

__ H20 + __ CO2 __ C6H12O6 + __ 02

REACTANTS PRODUCTS

STEP 1Write and balance the equation

__ H20 + __ CO2 __ C6H12O6 + __ 02

REACTANTS PRODUCTS

Water + Carbon Dioxide (+ energy) = Glucose + Oxygen

STEP 2Start with one of the knowns (convert mg to g)

6720 mg of H20 (Given)

STEP 2Start with one of the knowns (convert mg to g)

6720 mg of H20 (Given)

6720 mg H20 1 gram H20

1000 milligrams H20

1 gram H20 = 1000 mg H20

STEP 3Convert grams to moles

STEP 3Convert grams to moles

6.72 g H20 1 mole H20

18 grams H20

1 mole H20 = 18 g H20

STEP 4Convert mole to moles

6 H20 + 6 CO2 C6H12O6 + 6 02

STEP 4Convert mole to moles

STEP 4Convert mole to moles

6 H20 + 6 CO2 C6H12O6 + 6 02

.373 mole H20 1 mole C6H12O6

6 mole H20

6 mole H20 = 1 mole C6H1206

STEP 5Convert moles to grams

.062 mole C6H1206 180 grams C6H1206

1 mole C6H1206

STEP 5Convert moles to grams

1 mole C6H1206 = 180 g C6H1206

You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6

Now figure out: 10.67 L CO2 = ? grams C6H12O6

STEP 1Convert L at STP to moles

10.67 L of CO2 (Given)

STEP 1Convert L at STP to moles

10.67 L of CO2 (Given)

10.67 L CO2 1 mole CO2

22.4 Liters CO2

22.4 L CO2 =1 mole CO2

STEP 2Convert moles to moles

6 H20 + 6 CO2 C6H12O6 + 6 02

STEP 2Convert moles to moles

6 H20 + 6 CO2 C6H12O6 + 6 02

.476 mole CO2 1 mole C6H12O6

6 mole CO2

6 mole CO2 = 1 mole C6H1206

STEP 3Convert moles to grams

STEP 3Convert moles to grams

.079 mole C6H12O6 180 grams C6H12O6

1 mole C6H12O6

1 mole C6H1206 = 180 g C6H1206

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)

CO2 is the excess reactant

How much excess CO2 ?(In grams)

14.22 grams – 11.16 grams =

You have figured out that: 6720 mg of H20 = 11.16 grams C6H12O6

10.67 L of CO2 = 14.22 mol C6H12O6

3.06 grams CO2 in excess

Find the number of molecules of glucose that precipitated.

What’s Next?

STEP 1Convert moles to molecules

0.62 mole of C6H12O6 (Found)

STEP 1Convert moles to molecules

0.62 mole of C6H12O6 (Found)

0.62 moles C6H12O6 6.02 x 1023 C6H12O6

1 mole C6H12O6

1 mole C6H1206 = 6.02 x 1023 molecules C6H1206

Find the number of molecules of glucose that precipitated.

RESTATING THE QUESTION:

Find the number of molecules of glucose that precipitated.

3.73 E22 molecules C6H12O6

RESTATING THE QUESTION:

THEORETICAL & PERCENT YIELDFind theoretical percent yield of C6H12O6

(Actual amount of Glucose obtained was 10.22 as stated before)

THEORETICAL & PERCENT YIELD

Theoretical Yield: (Already Found) 11.16 grams

Find theoretical percent yield of C6H12O6

(Actual amount of Glucose obtained was 10.22 as stated before)

THEORETICAL & PERCENT YIELD

Theoretical Yield: (Already Found) 11.16 grams

Percent Yield: ACTUAL YIELD/THEORETICAL x 100

Find theoretical percent yield of C6H12O6

(Actual amount of Glucose obtained was 10.22 as stated before)

THEORETICAL & PERCENT YIELDFind 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%