Limiting Reactant & Percent Yield Notes

1. Limiting Reactant & Percent Yield Notes

2. Background Knowledge Check Label the reactant(s) and product(s) in the following reaction: 2 Mg + O2 2MgO Reactant(s): Product(s): Mg and O2 MgO

3. Limiting Reactant Manufacturers of cars and bicycles order parts in the same proportion as they are used in their product. Car manufacturers order four times as many wheels as engines and bicycle manufacturers order twice as many pedals as seats. In the same manner, when chemicals are mixed together so they can undergo a reaction, they are often mixed in stoichiometric quantities – exactly the correct amounts so that all the reactants “run out” at the same time. If the chemicals aren’t mixed to run out at the same time, one of the chemicals will limit or halt the reaction from taking place any further. The reactant that “runs out” or limits the reaction is called the limiting reactant. The reactant that still remains or is extra is called the excess reactant. In any stoichiometric problem, where reactants are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting in order to calculate correctly the amounts of products that will be formed.

4. Analogy: Baking Cookies A recipe calls for 2 cups of flour for every egg. You have 5 cups of flour and 3 eggs. What is your limiting ingredient? What is your excess ingredient? Steps for Solving Stoichiometry Problems Involving Limiting Reactants • Write and balance the equation for the reaction, if necessary. • For each reactant, convert grams reactant to grams product. • Compare grams of product: • The smaller grams of product is the theoretical yield and is the amount of product made • The smaller grams of product came from the limiting reactant • The larger grams of product came from the excess reactant flour eggs

5. Mg 1 x 24.3 = 24.3 O 1 x 16.0 = 16.0 40.3 g/mol Ex: 7.24 moles of Mg and 3.86 moles of O2 react to form MgO. 2 Mg + O2 2MgO How many grams of MgO are formed ? What is the limiting reactant ? What is the excess reactant ? Mg O2 7.24 mol Mg 2 mol MgO 40.3 g MgO x ______________ x ____________ 292 g MgO = 2 mol Mg 1 mol MgO 2 mol MgO 3.86 mol O2 40.3 g MgO x _____________ 311 g MgO x ____________ = 1 mol O2 1 mol MgO

6. N 1 x 14.0 = 14.0 H 3 x 1.0 = 3.0 17.0 g/mol Ex: Suppose 2.50 x 104 g of N2 and 5.00 x 103 g of H2 are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when the reaction is run to completion. N2 + 3 H2 2 NH3 What is the limiting reactant? What is the excess reactant? H2 N2 2.50 x 104 g N2 1 mol N2 mol NH3 17.0 g NH3 2 x ___________ x ______________ x ___________ 28.0 g N2 mol NH3 1 mol N2 1 = 3.04 x 104 g NH3 17.0 g NH3 5.00 x 103 g H2 1 mol H2 2 mol NH3 x ______________ x ___________ x ___________ mol NH3 2.0 g H2 3 mol H2 1 = 2.8 x 104 g NH3 N 2 x 14.0 = 28.0 g/mol H 2 x 1.0 = 2.0 g/mol

7. Percent Yield Theoretical yield – amount of product predicted from the amounts of reactants used, calculated from the limiting reactant Actual yield – amount of product actually obtained through experiment Percent yield – comparison of actual and theoretical yield Percent Yield = Actual yield X 100 Theoretical yield

8. C 1 x 12.0 = 12.0 H 4 x 1.0 = 4.0 O 1 x 16.0 = 16.0 32.0 g/mol Example: Methanol, CH3OH, can be produced by the reaction between carbon monoxide and hydrogen. Suppose 6.85 x 104 g of CO is reacted with 8.60 x 103 g of hydrogen. CO + 2H2 CH3OH • Calculate the theoretical yield of methanol. • If 3.57 x 104 g of CH3OH is actually produced, what is the percent yield of methanol? 6.9 x 104 g CH3OH 6.85 x 104 g CO 1 mol CO 1 mol CH3OH 32.0 g CH3OH x ___________ x ______________ x ___________ 28.0 g CO 1 mol CH3OH 1 mol CO 7.83 x 104 g CH3OH = 8.60 x 103 g H2 1 mol H2 1 32.0 g CH3OH mol CH3OH x ___________ x ___________ x ______________ 2.0 g H2 2 mol CH3OH mol H2 1 H 2 x 1.0 = 2.0 g/mol 6.9 x 104 g CH3OH = C 1 x 12.0 = 12.0 O 1 x 16.0 = 16.0 28.0g/mol 3.57 x 104 g 52 % X 100 = % Yield = 6.9 x 104 g