2Limiting ReactantsRarely are the reactants in a chemical equation present in the exact ratios specified by the balanced equation. Generally, there is too much of one and not enough of the other. The chemical reaction will stop once one of the reactants is used up. This leaves the remainder of the reactant that was in excess unreacted. The chemical that you ran out of was the limiting reactant.
3If we have… how many sets can we make? Limiting reactant = the reactant that runs out first causing the reaction to stopExcess reactant = reactant that does not get completely used up during a reactionWith this left overExcess
4Limiting Factor NoteSheet Consider the following recipe:3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder 12 muffinsA quick check of the pantry shows that the following quantities are available. How many muffins could be made with respect to each ingredient?9 cups of flour could make ___36___ muffins4 eggs could make __24__ muffins8 cups of sugar could make __96___ muffins10 tsp baking powder could make __60___ muffins
5Limiting Factor NoteSheet 3 cups flour + 2 eggs + 1 cup sugar + 2 tsp baking powder 12 muffinsWhat is the maximum number of muffins that could be made considering ALL ingredients? __24___Which ingredient would you run out of first? _eggs____This ingredient would be referred to as the __limiting_ __reactant____.The other ingredients are said to be __in __ excess .When you run out of an ingredient, you stop producing the _product__.
6Limiting Factor Notesheet Consider the following “recipe” (equation): 2 mol of Al + 3 mol of Cl2 2 mol of AlCl3 A quick check of the chemistry lab shows that the following quantities are available. How much product could be made from each? (How many times could you make the recipe?) 10 moles Al ______ 12 moles Cl2 _____
7What is the maximum amount of AlCl3 that could be made considering all the ingredients? ___8 moles_____We would say the aluminum is ___in excess______________ and the chlorine is _______limiting__________.What if you were presented with grams of each reactant? Let’s say there are 100 g of Al and 125 g of Cl2 available. How much product could be formed? (Remember, the coefficients in the equation are mole numbers)100 g of Al would yield ________ g of product.125 g of Cl2 would yield ________ g of product.How much product could be made considering both reactants? ________The limiting factor would be ________.
8Steps to solve limiting reactants problems Split into 2 problemsSolve each problemPick smallest answerSmallest answer comes from limiting reactantLargest answer comes from excess reactant
9Calculating the Product when the Reactant is Limited In the following equation:If 200 g of sulfur reacts with 100 g of chlorine what mass of sulfur chloride will be produced?S Cl2 4 S2Cl2Sulfur200 g S x 1 mole S x 4 mol S2Cl2 x 134 g S2Cl2 = 3350 g g S mole S mole S2Cl2Chlorine100 g Cl2 x 1 mole Cl2 x 4 mol S2Cl2 x 134 g S2Cl2 = g lowest #70 g Cl mol Cl mole S2Cl2
10Steps for Calculating Limiting Reactants 1. Balance the equation2. Label the known substances and the unknown substance for each reactant (note: you will have 2 problems)3. Convert from mass to moles for each reactant4. Use a mole ratio for each of the known substances for each reactant5. Use the molar mass to convert from moles to mass for each reactant6. Look for the lowest number – this is your limiting reactant
11Calculating product when reactant is limiting If we had 12 moles of nitrogen and 18 moles of hydrogen, what is the maximum number of moles of NH3 that could be produced?N2 + H2 NH3
12Calculating product when reactant is limiting If we had 112 grams of nitrogen and 18 grams of hydrogen, what is the maximum number of grams of NH3 that could be produced?N2 + 3H2 2 NH32(14) (1)NitrogenPick lowest numberTherefore hydrogen is your limiting reactantHydrogen
13Calculating product when reactant is limiting With 48 grams of magnesium and 48 grams of oxygen available, how much product can be formed?2Mg + O2 2 MgO(16)MagnesiumPick lowest numberTherefore magnesium is your limiting reactant and oxygen is the excess reactantOxygen
14Limiting Reactants Practice How many grams of sodium chloride can be produced in the following reaction with 150 grams of sodium and 200 grams of chlorine?2 Na + Cl2 2 NaCl(35)SodiumChlorinePick lowest number
15Limiting Reactants Practice In the above reaction, how many more grams of chlorine will be required to fully react with the remaining sodium?We need to figure out how many grams of chlorine is needed to react with 150 grams of sodium2 Na + Cl2 2 NaCl(35)Chlorine needed = gChlorine given (from problem) = 200 gAdditional Chlorine needed = gSodium
16Percent YieldActual yield = the actual amount of product formed during an experimentTheoretical yield = amount of product that could be produced according to calculationsPercent yield = ratio of actual and theoretical yield
17Percent Yield Practice A student uses stoichiometry to calculate the predicted yield of CO2 in a chemical reaction as 323 g. When the student mixes the chemicals to make the CO2, he finds that the reaction only produces 308 g of CO2. What is the percent yield?
18In the reaction below, how many grams of potassium oxide can be produced with 78 grams of potassium and 140 grams of boron oxide?6 K + B2O3 3 K2O + 2 B(11)+3(16) (39)PotassiumPick lower number L.R. = theoretical yieldBoron Oxide
19Theoretical yield = 94 g (from last slide) Problem 2 continuedIf the actual yield from the above reaction was 81 grams, what is the percent yield?Actual yield = 81 gramsTheoretical yield = 94 g (from last slide)Round your final answer to the nearest hundredths!
203. The actual yield in the above reaction was 39 grams of ammonium 3. The actual yield in the above reaction was 39 grams of ammonium. Using 84 grams of nitrogen and 8 grams of hydrogen, what is the percent yield?-first we must calculate the theoretical yield of product which means we need to find our limiting reactantN2 + 3 H2 2 NH32(14) (1)NitrogenPick lower numberL.R. = theoretical yieldHydrogen
21Problem 3 continuedActual yield = 39 gramsTheoretical yield = g (from last slide)