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

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?

MASS - MASS STOICHIOMETRY STEP 2 - BALANCE EQUATION 4 NH3 + 3 O2 ----> 2 N2 + 6 H2O

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

MASS - MASS STOICHIOMETRY STEP 3 - 4 NH3 + 3 O2 ----> 2 N2 + 6 H2O 34 grams 34 grams x 1mole = 17 grams Molar mass of ammonia 2 moles of ammonia

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

MASS - MASS STOICHIOMETRY STEP 4 - 4 NH3 + 3 O2 ----> 2 N2 + 6 H2O 2 moles 4 moles of NH3 = 2 moles of NH3 6 moles of water X X = 3 moles of water

MASS - MASS STOICHIOMETRY STEP 5 - Convert moles of “B” into mass of “B” by multiplying by the MOLAR MASS of “B” 4 NH3 + 3 O2 ----> 2 N2 + 6 H2O 3 moles 3 moles x 18 grams = 1 mole Molar mass of water 54 grams of water

MASS - MASS STOICHIOMETRY NOW LETS SEE THE WHOLE THING AT ONCE… 4 NH3 + 3 O2 ----> 2 N2 + 6 H2O 34 grams 2 moles 3 moles 54 grams

The Question...

The Answer