Stoichiometry & Limiting Reactants

Stoichiometry & Limiting Reactants

Name ______

Stoichiometry & Limiting Reactants

Textbook: 4.2

Stoichiometry

The mathematical method used to determine the amount of productsformed from a given set of reactants.

Eg. 1Write the equation for Aluminum reacts with oxygen gas to produce aluminum oxide. Check it with MsO before moving on.

a) How many moles of O2 are needed to produce 2 mole of aluminum oxide?

b) How many moles of Al are needed to produce 1 mole of aluminum oxide?

c) How many moles of O2 are needed to produce 6 mole of Al2O3?

d) If you have 1.52 mol O2 how many mol of product can be made?

e) If you have 0.25 mol Al, how many mol of product can be made?

Eg. 2Using the equation in Eg. 1, how many grams of Al are needed to produce 2 mole of Al2O3?

Practice: Tro Ch4 25, 27, 29

Eg. 3In the space below write out the equation from Eg. 1.

a) Determine the mass of Aluminum oxide produced if 42.5g Al are used.

b) Determine the mass of O2 needed for the reaction.

Eg. 4Zinc reacts with hydrochloric acid to produce hydrogen gas and zinc chloride.

a) Write the balanced chemical equation for the reaction. Check it with MsO before moving on.

b) If 42.5 g of hydrochloric acid (MM=36.453g) is reacted, calculate the mass (theoretical yield) of zinc chloride

(MM= 136.296g) will be produced.

c)What mass of Zn has it reacted with?

Eg. 5Imagine that you are a cheap chemistry teacher who is trying to recover silver from a solution used in lab. The

redox rxn between copper & silver nitrate produces metallic silver & copper (II) nitrate. Suppose the reaction is

not worth the trouble unless you yield 500 g or more of silver. What mass of copper & silver nitrate would you

need to react together to produce 500 g of silver?

Practice: Tro Ch4 CC 4.1 & 4.2, Q1, and 31, 33, 35, 32

Before proceeding complete the stoichiometry “cards” activity.

Textbook: 4.3Important terms: pg 146

Limiting reactants

Until now we have assumed that we do not have any excess reactant in a reaction. Typically this is not the case.

In any reaction one of the reactants limits or restricts the amount of product that can be made (the limiting reactant) the other reactant (s) are said to be in excess, that is, some remains after the reaction has completed.

The maximum predicted mass (theoretical yield) for any reaction is based on the limiting reactant.

Eg. 6In the space below write the equation from Eg. 1.

a) If you have 14Al atoms & 12 O2 molecules, how many Al2O3mlcls can you make?

Which reactant limits the production of aluminium oxide?

How much do you have of the excess?

b) If you have 9Al atoms & 6O2 molecules, how many Al2O3 can you make?

Which reactant limits the production of aluminium oxide?

How much do you have of the excess?

Write the equation:

c) If you have 1.21 molAl & 2.65 mol O2 molecules, how many mol Al2O3 can you make?

Practice: Tro Ch4 CC4.3, Q and 37, 39, 40, 41, 43

d) If you have 79.1g Al atoms & 80.1 g O2 molecules, calculate the theoretical yield, in grams, forAl2O3 ?

How much of the excess reactant remains?

Eg. 7Suppose 10.45 g of aluminum metal are reacted with 66.55 g of copper (II) sulfate (mm=159.6) in a redox rxnyielding aluminum sulfate (mm=330.23g) and copper metal as products.

a) Which reactant is in excess?

Write the rxn:

b) Calculate the mass of the excess.

c) Calculate the theoretical yield, in grams, for each product

Practice: Tro Ch4 CC 4.4, Q2, Q3 and 45, 47, 48, 49, 51, Q4
Percent yield

Theoretical yield amount of product predicted using stoichiometry based on the limiting reactant.

Actual yield – amount of product obtained in the lab from the reaction in an experimental setting.

% yield = actual x 100

theoretical

(calculated)

Eg. 8 Using the information in eg. 7, calculate the percent yield if 23.27g of Cu were isolated at the end of an

experiment.

Additional Zumdahl practice: Ch.358a,b 65a,b,c 99 101 102 109a 111 114 113

Even answers: 58a) 70.1g Nb) 1.40x102 g N

102a) Ba(OH)2*8H2O + 2NH4SCN  Ba(SCN)2 + 10H2O + 2NH3

102b) 3.2 g NH4SCN

114a) 1170 g C14H9Cl5b) C2HOCl3 is limitingc) 401g C6H5CL in excessd) 17.1%