Limiting Reactant There are MANY different ways to solve Limiting Reactant questions depending on what is asked – this is one way…

What was the “limiting reactant?” How many BLT Sandwiches can you make with 3 Tomatoes, 2 Kg of Bacon, 1 head of lettuce and 4 slices of bread? 2 What was the “limiting reactant?” How did you identify it?

OUTCOME QUESTION(S): C11-3-15 STOICHIOMETRY Excess reactant Solve stoichiometric problems involving moles, mass, and volume, given the reactants and products in a balanced chemical reaction. Include: heat of reaction Identify the limiting reactant and calculate the mass of a product, given reaction equation and reactant data. Include: theoretical yield, experimental yield Vocabulary & Concepts  Excess reactant

No more products can be made, regardless of “extra” nitrogen Limiting Reactant - determines the amount of product that can be formed in a reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) 2 moles + 3 moles N H H N H N N H H No more products can be made, regardless of “extra” nitrogen Reactants remaining are called the excess reactants.

How to Solve Limiting Reactant Problems Step 1: Record what you HAVE Step 2: Calculate what you will USE Pick one reactant and calculate how much of the other you will use. Step 3: Identify the limiting reactant Ask “Do I have enough?” Step 4: Use limiting reactant to determine the amount of product Don’t panic! The calculations are just more of the same stoichiometry

Pick EITHER reactant and calculate the other How much NaCl is produced when 6.70 mol of Na react with 3.20 mol of Cl2? 6.70 mol 3.20 mol 2 Na (s) + Cl2 (g) 2 NaCl (s) HAVE USE Pick EITHER reactant and calculate the other 6.70 mol Na 1 Cl2 = 3.35 mol 3.35 mol Cl2 (use) 2 Na Step 1: Record what you HAVE Step 2: Calculate what you will USE Pick one reactant and calculate how much of the other you will need.

Cl2 is the Limiting Reactant For comparison: let’s pretend we picked the other reactant… 2 Na (s) + Cl2 (g) 2 NaCl (s) HAVE 6.70 mol 3.20 mol USE 3.35 mol 3.20 mol Cl2 2 Na = 6.40 mol 6.40 mol Na (use) 1 Cl2 Both calculations lead to the SAME conclusion: Too much Na OR Not enough Cl2 Cl2 is the Limiting Reactant Step 3: Identify the limiting reactant Ask “Do I have enough?”

Na - excess reactant Cl2 - limiting reactant 2 Na (s) + Cl2 (g) 2 NaCl (s) ? HAVE 6.70 mol 3.20 mol USE 6.40 mol 3.35 mol 3.20 mol Cl2 2 NaCl = 6.40 mol NaCl (made) 1 Cl2 So this reaction will use ALL 3.20 moles of Cl2 and only 6.40 moles of Na to make 6.40 moles of NaCl This is the step most mess up on – be sure to start with the right value (what you HAVE of the Limiting Reactant) Step 4: Use limiting reactant to determine the amount of product

Pick EITHER reactant and calculate the other How much CO2 is produced with 10 mol of propane and 10 mol of O2 and what is excess is left over? C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 HAVE 10 mol 10 mol USE At this point we know the LR (we don’t HAVE enough)  Oxygen We still don’t know how much excess C3H8 there is… Pick EITHER reactant and calculate the other 10 mol C3H8 5 O2 = 50 mol 50 mol O2 (use) 1 C3H8 Step 1: Record what you HAVE Step 2: Calculate what you will USE Pick one reactant and calculate how much of the other you will need. Step 3: Identify the limiting reactant Ask “Do I have enough?”

C3H8 - excess reactant O2 - limiting reactant C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 HAVE 10 mol 10 mol - USE 50 mol ? 8 mol (excess) We have to go the “other way” too – to find what is left over 10 mol O2 1 C3H8 = 2 mol 2 mol C3H8 (use) 5 O2 NOTE: Finding the excess left over requires the HAVE and USE for the excess reactant – this might have already been done IF you were lucky enough to pick the right reactant to start with in step 2…

C3H8 - excess reactant O2 - limiting reactant C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 ? HAVE 10 mol 10 mol - USE 2 mol 50 mol 8 mol (excess) 10 mol O2 3 CO2 = 6 mol CO2 (made) 5 O2 So this reaction will use ALL 10 moles of O2 and only 2 moles of C3H8 to make 6 moles of CO2 Unless asked, the other “USE” value – C3H8 - isn’t necessary to finding the product, only for finding the excess Step 4: Use limiting reactant to determine the amount of product

3 2 N2 (g) + H2 (g) NH3 (g) = 16.3 g N2 (use) = 3.90 g H2 (use) How many grams of ammonia are made from 3.50 g of H2 gas and 18.0 g of nitrogen gas? What’s left? N2 (g) + H2 (g) NH3 (g) 3 2 HAVE 18.0 g 3.50 g N2 = 28.0 g/mol USE 16.3 g 3.90 g H2 = 2.0 g/mol 3.50 g H2 1 mol H2 1 N2 28.0 g N2 = 16.3 g N2 (use) 2.0 g H2 3 H2 1 mole N2 Did you “pick” right? IF so, only one of these calculations was necessary! 18.0 g N2 1 mol N2 3 H2 2.0 g H2 = 3.90 g H2 (use) 1 mole H2 28.0 g N2 1 N2

N2 - excess reactant H2 - limiting reactant N2 (g) + H2 (g) NH3 (g) 3 2 HAVE 18.0 g 3.50 g ? - USE 16.3 g 3.90 g 1.7 g (excess) N2 - excess reactant H2 - limiting reactant NH3 = 17.0 g/mol 3.50 g H2 1 mol H2 2 NH3 17.0 g NH3 = 19.8 g NH3 (made) 2.0 g H2 3 H2 1 mole NH3 Step 3: Identify the limiting reactant Ask “Do I have enough?” Step 4: Use limiting reactant to determine the amount of product

5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g) What volume of CO2 is formed by the reaction of 75.0 g of propane and 150.0 L of oxygen at STP? C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 HAVE 75.0 g 150.0 L USE 58.9 g 191 L C3H3 = 44.0 g/mol 75 g C3H8 1 mol C3H8 5 mol O2 22.4 L O2 = 191 L O2 44 g C3H8 1 mol C3H8 1 mol O2 Did you “pick” right? 150 L O2 1 mol O2 1 mol C3H8 44 g C3H8 = 58.9 g C3H8 22.4 L O2 5 mol O2 1 mol C3H8

- 5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g) ? HAVE 75.0 g 150.0 L - USE 58.9 g 191 L 16.1 g (excess) C3H8 - excess reactant O2 - limiting reactant 150 L O2 1 mol O2 3 mol CO2 22.4 L CO2 = 90 L CO2 22.4 L O2 5 mol O2 1 mol CO2 So this reaction will use ALL 150 L of O2 and only 58.9 g of C3H8 to make 90 L of CO2 Step 3: Identify the limiting reactant Ask “Do I have enough?” Step 4: Use limiting reactant to determine the amount of product

C11-3-15 STOICHIOMETRY Excess reactant CAN YOU / HAVE YOU? Solve stoichiometric problems involving moles, mass, and volume, given the reactants and products in a balanced chemical reaction. Include: heat of reaction Identify the limiting reactant and calculate the mass of a product, given reaction equation and reactant data. Include: theoretical yield, experimental yield Vocabulary & Concepts  Excess reactant