Sec 12.3 limiting reactant, percent, actual and theoretical Yield

Excess Reactant- the reactant that is leftover in a chemical reaction. Definitions: Limiting Reactant- the reactant that gets used up in a chemical reaction. Excess Reactant- the reactant that is leftover in a chemical reaction. Analogy: When making brownies you have plenty of flour, but run out of eggs. The flour is the ___________ reactant, and the eggs are the _____________ reactant.

Excess Reactant- the reactant that is leftover in a chemical reaction. Definitions: Limiting Reactant- the reactant that gets used up in a chemical reaction. Excess Reactant- the reactant that is leftover in a chemical reaction. Analogy: When making brownies you have plenty of flour, but run out of eggs. The flour is the excess reactant, and the eggs are the limiting reactant.

To calculate a Limiting Reactant: 1. Convert grams of each reactant to grams of one of the products. 2. Compare the grams of product for each. Whichever gives LESS product is the limiting reactant.

Ex. Which reactant is limiting if 25. 0g of N2 and 5 Ex. Which reactant is limiting if 25.0g of N2 and 5.0 g of H2 are mixed together? N2 + H2  NH3 Convert g mol  mol  g for each reactant Balanced equation:

Ex. Which reactant is limiting if 25. 0g of N2 and 5 Ex. Which reactant is limiting if 25.0g of N2 and 5.0 g of H2 are mixed together? N2 + H2  NH3 Convert g mol  mol  g for each reactant Balanced equation: N2 + 3H2  2NH3

N2 + 3H2  2NH3 25.0 gN2- g N2 mol N2  mol NH3  g NH3 25.0 g N2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 28.02 g N2 1 mol N2 1 mol NH3 = 30.4 g NH3

N2 + 3H2  2NH3 = 28 g NH3 5.0 g H2- g H2 mol H2  mol NH3  g NH3 5.0 g H2 x 1 mol H2 x 2 mol NH3 x 17.04 g NH3 2.02 g H2 3 mol H2 1 mol NH3 = 28 g NH3

2. Compare the two amounts of NH3 Starting with 25.0 g N2 = 30.4 g NH3 Starting with 5.0 g H2 = 28 g NH3 limiting: excess.

2. Compare the two amounts of NH3 Starting with 25.0 g N2 = 30.4 g NH3 Starting with 5.0 g H2 = 28 g NH3 limiting: H2 excess: N2

Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu(s) + S(s)  Cu2S(s) What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

Theoretical Yield The maximum amount of product that can be produced. It is rarely obtained because of experimental errors

Actual (experimental) Yield This is the amount of product that is actually obtained.

Percent Yield

Ex 1: What is the percent yield if the theoretical yield of product is 45g , but when doing the experiment you only got 30g of product?

Ex 1: What is the percent yield if the theoretical yield of product is 45g , but when doing the experiment you only got 30g of product? Actual Yield= 30 g Theoretical Yield = 45 g Percent Yield = (30g / 45g) x 100 = 66.67%

Ex 2: You perform a synthesis reaction to make water from oxygen and hydrogen. You expect to produce 25,000 grams of water, but only get 18,750 g H2O what is your percent yield?

Ex 2: You perform a synthesis reaction to make water from oxygen and hydrogen. You expect to produce 25,000 grams of water, but only get 18,750 g H2O what is your percent yield? Actual Yield= 18,750 g Theoretical Yield = 25,000g Percent Yield = ? X = 18,750 g x 100 = 75% 25,000 g

Al (s) + I2 (s)  Al I3 (s) 2Al (s) + 3I2 (s)  2AlI3 (s) Ex 3. Al (s) + I2 (s)  Al I3 (s) If we react 2.40 g I2, but only produce 2.05 g Al I3, what is the percent yield? Balance the equation: Find out the Theoretical Yield of Al I3 Find the Percent Yield Balance the Equation Al (s) + I2 (s)  Al I3 (s) 2Al (s) + 3I2 (s)  2AlI3 (s)

Find the Theoretical Yield (yes, three step conversion) g I2  mol I2 mol AlI3 g AlI3 2.40 g I2 x 1 mol I2 x 2 mol AlI3 x 407.7g AlI3 253.8 g I2 3 mol I2 1 mol AlI3 = 2.57g AlI3

Find the Percent Yield Actual Yield= 2.05 g Theoretical Yield = 2.57 g (2.05g / 2.57g) x 100% = 79.77 %