Unit+4-Limiting+Reagents

-This section is a specific part of Stoichiometry that poses the question: "When you mix things together, is there anything left over?" -The purpose is to determine which reactant is left over(in excess), and which reactant was used up completely(limiting reagent). -A limiting reagent is the reactant which is completely used up and determines when the reaction will stop; this can be determined stoichiometrically.

__Ex:__ If you had 3 moles of Nitrogen and 3 moles of hydrogen, how much Ammonia could you create? Which would be the limiting reagent and which is the excess reagent? -Write out the balanced equation. N2 + 3H2 -> 2NH3 -The mole ratio tells us that 1 mol nitrogen reacts 3 mol hydrogen. -Looking back at our problem, we realize that the 3 mol hydrogen limits the reaction, because it will react with 1 mol of nitrogen, leaving 2 moles Nitrogen in excess.

__Ex:__ Now given the combustion of propane: C3H8 + 5O2 -> 3CO2 + 4H2O -If you had 20.0mol propane and 25 mol of oxygen, which is the limiting reagent and which is the excess reagent? __Therefore:__ The propane is in excess and the oxygen is the limiting reagent due to the combustion ratio of 5:1.

__Ex:__ If 200.0g of Sulfur reacts with 100.0g of Chlorine, what mass of Disulfur dichloride is produced? How much Sulfuris leftover?

__Step 1:__ Write the balanced equation. S8 + 4Cl2 -> 4S2Cl2

__Step 2:__ Determine the limiting reagent. -Convert to mole. 200g Sulfur/256.52 = 0.7797mol S8(subscript needed) 100g Chlorine/70.906 = 1.410mol Cl2 -Unless you have already determined which is the limiting reagent, proceed to the next step.

__Step 3:__ Compare your results to the mole ratio in the balanced equation. -Balanced equation -> 1mol S8/4mol Cl2 -Our given data -> 0.7797mol S8/1.410mol Cl2 -Reduce this ratio!(/ both by lowest number) -This yields: 1mol S8/1.809 mol Cl2 -Compare these ratios: -The balanced ratio says that you need 1:4, the given data yields a ratio of 1:1.809. -There isn't enough chlorine so our limiting reagent is __Chlorine__. -This is caused by the ratio only being 1:1.809 rather then 1:4. -The limiting reagent determines how much of the other reactant is used, and how much product is obtained.

__Step 4:__ Determine the amount of disulfur dichloride produced. -1.410mol Cl2 x 4mol S2Cl2/4mol Cl2 = __1.410 mol S2Cl2

Step 5:__ Determine how much S8 is leftover. -1.410mol Cl2 x 1mol S8/4mol Cl2 = __0.3525mol S8__ -0.7797mol S8(started with) - 0.3525mol S8(used) = __0.4272 mol S8(leftover)

Example 2:__ 25.0g of Carbon burns in the presence of 50.0g of oxygen, creating Carbon dioxide. a) __Balanced equation:__ C + O2 -> CO2 b) __Limiting Reagent:__ 25.0g Carbon = 2.08mol C 50.0g Oxygen = 1.56mol O2 -Therefore Oxygen is the limiting reagent since there is less of it, and it is a 1:1 ratio. c) __Mass of excess:__ 1.56mol O2 x 1mol C/1mol O2 = 1.56mol C x 12.01g C/1mol C = __18.7g C__ 25.0g C(started with) - 18.7g C(used) = __6.3g (end)__ d) __Mass Product:__ 1.56mol O2 x 1mol CO2/1mol O2 x 44.01g CO2/1mol CO2 = __68.7g CO2__

Here are some helpful links for extra practice and explanation on Limiting Reagents: http://www.chempractice.com/drills/java_Limiting.php http://scidiv.bcc.ctc.edu/bg/limiting.html http://www.youtube.com/watch?v=FzU0udv8PA4