2Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data.Include: theoretical yield, experimental yieldAdditional KEY TermsExcess reactant
3How many BLT Sandwiches can you make with 3 Tomatoes, 2 Kg of Bacon, 1 head of lettuce and 4 slices of bread?2
4Limiting Reactant - determines the amount of product that can be formed in a reaction. N2 (g) H2 (g)2 NH3 (g)2 moles molesNHHNHNNHHReactants remaining are called the excess reactants.
5Limiting Reactant Problems Step 1: Record what you HAVEStep 2: Calculate what you NEEDPick one reactant and calculate how much of the other you will need.Step 3: Determine the limiting reactant and excess reactantStep 4: Use limiting reactant to determine the amount of product.
6Pick one reactant and calculate the other How much NaCl is produced and what is left over when 6.70 mol of Na react with 3.20 mol of Cl2?6.70 mol3.20 mol2 Na (s) + Cl2 (g) NaCl (s)HAVENEEDPick one reactant and calculate the other6.70 mol Na1 mol Cl2=3.35 molCl2 (need)2 mol Na
72 Na (s) + Cl2 (g) 2 NaCl (s) = 6.40 mol Na (need) 3.20 mol Cl2 HAVE mol molNEED molPick one reactant and calculate the other3.20 mol Cl22 mol Na=6.40 molNa (need)1 mol Cl2Both calculations lead to the same conclusion:Cl2 - limiting reactant Na - excess reactant
8Cl2 - limiting reactant Na - excess reactant 2 Na (s) + Cl2 (g) NaCl (s)HAVE mol molNEED mol molCl2 - limiting reactant Na - excess reactant3.20 mol Cl22 mol NaCl= mol NaCl1 mol Cl2You could use your data to calculate exactly how much excess is left over:6.70 mol mol = 0.30 mol Na excess
9N2 (g) + 3 H2 (g) 2 NH3 (g) = 16.3 g N2 = 3.90 g H2 How many grams of ammonia can be made from 3.50 g of H2 gas and 18.0 g of nitrogen gas? What’s left?N2 (g) H2 (g) NH3 (g)HAVE g gNEED16.3 g3.90 g3.50 g H21 mol H21 mol N228.0 g N2= g N22.0 g H23 mol H21 mole N218.0 g N21 mol N23 mol H22.02 g H2= g H228.0 g N21 mol N21 mole H2
10H2 - limiting reactant N2 - excess reactant N2 (g) H2 (g) NH3 (g)HAVE g gNEED16.3 g3.90 gH2 - limiting reactant N2 - excess reactant3.50 g H21 mol H22 mol NH317.0 g NH32.0 g H23 mol H21 mol NH3= g NH318.0 g – 16.3 g = 1.70g N2 left
115 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g) What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and L of oxygen at STP?C3H8 (g) O2 (g) CO2 (g) + H2O (g)534HAVE g LNEED58.9 g191 L75 g C3H81 mol C3H85 mol O222.4 L O2= 191 L O244 g C3H81 mol C3H81 mol O2150 L O21 mol O21 mol C3H844 g C3H8= 58.9 g C3H822.4 L O25 mol O21 mol C3H8
12O2 - limiting reactant C3H8 - excess reactant C3H O CO H2OHAVE g LNEED58.9 g191 LO2 - limiting reactant C3H8 - excess reactant150 L O21 mol O23 mol CO222.4 L CO2= 90 L CO222.4 L O25 mol O21 mol CO275.0 g – 58.9 g = 16.1 g C3H8 left
13The limiting reactant is completely consumed. The excess reactant is NOT used up.When solving limiting reactant problems:Balance the chemical equation firstFind the limiting reactantUse limiting reactant to determine the productCalculate the excess
14CAN YOU / HAVE YOU?Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data.Include: theoretical yield, experimental yieldAdditional KEY TermsExcess reactant