1. MASS - MASS STOICHIOMETRY
Given the mass of a reactant in a reaction, the masses of the products can be determined.

2. MASS - MASS STOICHIOMETRY
PROBLEM: When 34 grams of ammonia reacts with enough oxygen, nitrogen and water vapor are yielded as products. How much water vapor is produced?

3. MASS - MASS STOICHIOMETRY
STEP 2 - BALANCE EQUATION
4 NH O > 2 N H2O

4. MASS - MASS STOICHIOMETRY
STEP 3 - Given the mass of “A”, calculate the moles of “A” by dividing by the MOLAR MASS of “A.”
GIVEN: 34 grams of ammonia

5. MASS - MASS STOICHIOMETRY
STEP 3 -
4 NH O > 2 N H2O
34 grams
34 grams x 1mole = grams Molar mass of ammonia
2 moles of ammonia

6. MASS - MASS STOICHIOMETRY
STEP 4 - Convert moles of “A” into moles of “B” using a molar ratio from the coefficients

7. MASS - MASS STOICHIOMETRY
STEP 4 -
4 NH O > 2 N H2O
2 moles
4 moles of NH3 = 2 moles of NH moles of water X
X = 3 moles of water

8. MASS - MASS STOICHIOMETRY
STEP 5 - Convert moles of “B” into mass of “B” by multiplying by the MOLAR MASS of “B”
4 NH O > 2 N H2O
3 moles
3 moles x 18 grams = mole Molar mass of water
54 grams of water

9. MASS - MASS STOICHIOMETRY
NOW LETS SEE THE WHOLE THING AT ONCE…
4 NH O > 2 N H2O
34 grams
2 moles
3 moles
54 grams

10. The Question...

11. The Answer