Follow the method Let the units guide you

When nitrogen and hydrogen react, they form ammonia gas, NH 3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced?

N 2 + H 2  NH 3 ? g NH 3 = 56.0 g N Use molar mass to convert grams to moles 2. Use mole ratio to switch from moles of N 2 to moles of NH g NH 3 3. Use molar mass to convert moles to grams

Go one step at time, using conversion factors until you have solved the problem

Sodium metal reacts with oxygen gas to produce solid sodium oxide. How many grams of sodium must react to produce 42.0 grams of sodium oxide?

Na(s) + O 2 (g)  Na 2 O(s) ? g Na = 42.0 g Na 2 O Use molar mass to convert grams to moles 2. Use mole ratio to switch from moles of Na 2 O to moles of Na 31.2 g Na 3. Use molar mass to convert moles to grams

When 12.0 gram of hydrogen reacts with oxygen, how many grams of water are produced?

H 2 (g) + O 2 (g)  H 2 O(l) ? g H 2 O = 12.0 g H Use molar mass to convert grams to moles 2. Use mole ratio to switch from moles of H 2 to moles of H 2 O g H 2 O 3. Use molar mass to convert moles to grams

Actual Yield Amount of product produced when the reaction is performed in a lab

Theoretical Yield Amount of product expected to be produced based on the balanced equation and the amount of reactants

Percent Yield (Actual / Theoretical Yield) x 100%

Percent Yield Lab In lab hydrochloric acid was added to sodium bicarbonate. A gas was formed (bubbles formed). After the reaction was complete, the remaining liquid was heated. NaHCO 3 + HCl  NaCl +CO 2 +H 2 O The gas bubbles away ↑

Percent Yield Lab In lab hydrochloric acid was added to sodium bicarbonate. A gas was formed (bubbles formed). After the reaction was complete, the remaining liquid was heated. NaHCO 2 + HCl  NaCl +CO 2 +H 2 O The gas bubbles away We then evaporate the water, leaving us with just NaCl.

Lab Data NaHCO g Mass before reaction5.98 g Mass after reaction 5.85 g Mass of NaCl produced0.13 g This amount it the actual yield.

NaHCO 3 + HCl  NaCl +CO 2 +H 2 O ? g NaCl = 0.23 g NaHCO 3 1. Use molar mass to convert grams to moles 2. Use mole ratio to switch from moles of Na 2 O to moles of Na 0.20 g Na 3. Use molar mass to convert moles to grams Now we will use stoichiometry to determine the theoretical yield

Compare Theoretical Yield to Actual Yield (Actual / Theoretical Yield) x 100% (0.13/0.16) x 100% = 81%

Propane C 3 H g CO g