ReactionStoichiometr

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Introduction

Reaction stoichiometry involves the mass relationshipsbetween reactants and products in a chemical reaction.

It is based on chemical equations similar to the ones studied in the last section.

All reaction stoichiometry calculations start with a balanced equation.

You will need to be familiar with gram/mole relationships as studied earlier this year.

C3H8 + 5O2 3CO2 + 4H2O

The mole enables chemists to move from the microscopic world of atoms and molecules to the real world of grams .

Stoichiometry problems are classified between the information given in the problem and the information you are expected to find, the unknown.

The given and the unknown may be expressed in grams or moles.

The masses in the reaction are usually expressed in grams.

Definition of mole:

mole of a substance = grams of substance/MW of substance

You will need to use: i. molar ratios in a balanced equation.ii. molar masses of reactants and products.iii. balancing equations.iv. conversions between grams and moles.

Mole RatiosA mole ratio converts moles of one

compound in a balanced chemical equation into moles of another compound.

All stoichiometry problems use mole ratios.

Example

Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks)

2 Mg(s) + O2(g) 2 MgO(s) (balanced)

Mole Ratios:2 : 1 : 2

Stoichiometry (working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)2 2H2O + BaCl2 1 1

2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

coefficients give MOLAR RATIOS

1) N2 + 3 H2 ---> 2 NH3

Write the mole ratios for N2 to H2 and NH3 to H2.

Practice Problems

Review: Molar Mass

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

CO2 = 44.01 grams per mole

H2O = 18.02 grams per mole

Ca(OH)2 = 74.10 grams per mole

Review: Chemical Equations

C2H5OH + 3O2 ® 2CO2 + 3H2O

reactants products

1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Types of Stoichiometry

Problems

Problem Type 1:

When you are given the amount of starting material in a reaction in moles and asked to calculate the amount of product in moles:

amount of reactant in moles amount of product in moles

Problem Type 2:

When you are given the amount of starting material in moles and asked to calculate the mass of product in grams:

amount of reactant in moles amount of product in moles mass of product in grams

Problem Type 3:

When you are given the mass of starting material in grams and asked to calculate the amount of product in moles.

amount of reactant in grams amount of reactant in moles amount of product in moles

Problem Type 4:

1. When you are given the mass of starting material in grams and asked to calculate the amount of product in grams.

amount of reactant amount of reactant in moles amount of product in moles amount of product

in grams in grams

Problem Type 1: Given and unknown quantities are in moles

Amount in molesof known substance

Amount in moles of unknown substance

CO2 + LiOH Li2CO3 + H2O

How many moles of lithium hydroxide are required to react with 20 moles of CO2

CO2 + 2 LiOH Li2CO3 + H2O balanced equation

Given: amount of CO2 = 20 moles

Unknown: amount of LiOH in moles

Amount of CO2 in moles Amount of LiOH in moles

mol CO2 x mol LiOH / mol CO2 = mol LiOH

20 mol CO2 x 2 mol LiOH / 1 mol CO2 = 40 mol LiOH

mole ratio

When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

2N2O5(g) 4NO2(g) + O2(g)

4.3 mol ? mol

Problem Type 1

Units match

Mole – Mole Conversions

When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5

52

2

ON mol2

NO mol48.6

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

Mole – Mole Conversions

Units match

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O2

4.3 mol N2O5

52

2

ON 2mol

O mol12.2

2N2O5(g) 4NO2(g) + O2(g)

4.3 mol ? mol

Problem Type 1: Given and unknown quantities are in moles

Amount in molesof known substance

Amount in moles of unknown substance

How many moles of ammonia are produced when 6 moles of hydrogen gas react with an excess of nitrogen gas.

3 H2 + N2 2 NH3 balanced equation

H2 + N2 NH3 unbalanced equation

Given: amount of H2 = 6 moles

Unknown: amount of NH3 in moles

Amount of H2 in moles Amount of NH3 in moles

(mol H2) x (mol NH3 / mol H2) = mol NH3

(6 mol H2) x (2 mol NH3 / 3 mol H2) = 4 mol NH3

Mole ratio

Problem Type 2: Given amount is in moles and unknown quantity is in grams

Amount in molesof known substance

Amount in grams of unknown substance

Problem Type 3: Given amount is in grams and unknown quantities are in moles

Amount in gramsof known substance

Amount in moles of unknown substance

Problem Type 2

mole ↔ gram

In plants when carbon dioxide reacts with water it produces glucose and oxygen:

6CO2 + 6H2O(l) C6H12O6(s) + 6O2(g)

How many grams of C6H12O6 is produced when 3.0 mol of water react with carbon dioxide?

= g C6H12O6

3.0 mol H2O

6126

6126

OHC mol1

OHC g2.18090

OH mol6

OHC mol1

2

6126

3.0 mol ? gramsUnits match

6CO2 + 6H2O C6H12O6 + 6O2

Problem Type 2

mole ↔ gram

In plants when carbon dioxide reacts with water it produces glucose and oxygen:

6CO2 + 6H2O(l) C6H12O6(s) + 6O2(g)

How many grams of CO2 is needed to react with 3.0 mol of water?

= g CO2

3.0 mol H2O

2

2

CO mol1

CO 4g4132

OH mol6

CO mol6

2

2

3.0 mol? gramsUnits match

6CO2 + 6H2O C6H12O6 + 6O2

Problem Type 2

mole ↔ gram

When magnesium burns in air, it combines with oxygento form magnesium oxide according to the following equation:

2Mg + O2(g) 2MgO(s)

How many grams of MgO is produced from 2.0 mol of magnesium?

= g MgO2.0 mol Mg

MgO mol1

MgO 0g480

Mg mol2

MgO mol2

2.0 mol ? gramsUnits match

2Mg + O2 2MgO

Problem Type 3

When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5

210 g NO2

2

52

NO mol4

ON mol22.28

2

2

NO g0.46

NO mol

gram ↔ mole

2N2O5(g) 4NO2(g) + O2(g)

210g? moles

Units match

Problem Type 3

Nitric acid is produced from the catalytic oxidation of ammonia

NH3(g) + O2(g) NO(g) + H2O(g)

a. How many moles of NO were produced from 824g of NH3?

= moles NO824 g NH3

3NH 4

NO mol4

mol48.4

3

3

NH 04.17

NH mol1

g

gram ↔ mole

824g ? moles

Units match

4NH3(g) + 5O2 4NO(g) + 6H2O(g)

Steps Involved in Solving Mass-Mass Stoichiometry Problems

• Balance the chemical equation correctly• Using the molar mass of the given substance,

convert the mass given to moles. • Determine the molar ratio.• Using the molar mass of the unknown

substance, convert the moles just calculated to mass.

Amount in gramsof known substance

Amount in molesof known substance

Amount in molesof unknown substance

Amount in gramsof unknown substance mw mwmolar

ratio

How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O5

75.0 g O2

2

52

O 1mol

ON mol2506

2

2

O g 32.0

O mol

52

52

ON mol

ON g108

2N2O5(g) 4NO2(g) + O2(g)75.0 g? grams

grams ↔ grams

When N2O5 is heated, it decomposes:

Problem Type 4

Stoichiometry Problem- Type 4

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.

1. Identify reactants and products and write the balanced equation.

Al + O2 Al2O3

a. What are the reactants?

b. What are the products?

c. What are the balanced coefficients?

Working a Stoichiometry Problem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =

12.3 g Al2O3

How many grams of acetylene are produced by adding water to 5.0 grams of CaC2?

= grams C2H2

5.0 g CaC2

2

22

CaC 1mol

HC mol 12.0

2

2

CaC g 64.1

CaC mol 122

22

HC mol 1

HC 26g

CaC2 + 2H2O C2H2 + Ca(OH)2 Balanced?

5.0 g ? grams

Acetylene gas (C2H2) is produced by adding water to calcium carbide:

How many moles of CaC2 are needed to react completelywith 49.0 g H2O:

CaC2 + 2H20 C2H2(g) + Ca(OH)2

How many moles of CaC2 are needed to react 49.0 g of H20?

= moles CaC2

49 g H2O

OH mol 2

CaC mol 1

2

21.36

OH 0.18

OH mol 1

2

2

g

CaC2 + 2H20 C2H2(g) + Ca(OH)2

49.0g? moles

Units match

Suppose you want to figure out how many grams of ammonia you can produce if you react 60.0 g of hydrogen gas with excess nitrogen.

Remember the balanced chemical equation gives the mole relationship between reactants and products.

N2 + 3 H2 2 NH3

1) Take the grams of hydrogen gas (recall that it is diatomic) and convert it to moles:

2) Convert the moles of hydrogen gas to moles of ammonia, using the coefficients in the balanced chemical equation:

3) Convert from moles of ammonia to grams of ammonia using the molecular mass and solve the equation:

Stoichiometry Review Problem

Calculating the Percent Yield

The predicted amount of product – which we have been doing with our stoichiometry problems has been for 100% yield, or the theoretical yield of a reaction.

The theoretical yield is the maximum amount of product that can be produces in a given reaction.

When chemical reactions take place, they are almost never 100% complete. A reaction may not go to 100% due to not all the reactants becoming involved, impurities in the reactants, competing side reactions, loss of product due to filtering, or just not getting it all out of the vessel.

The actual yield is how much product can be collected – and measured. The percent yield is a ratio given by:

Percent Yield = Actual Yield x 100% Theoretical Yield

Calcium carbonate is decomposed by heating, as shown in the following equation:

CaCO3 CaO + CO2

1) What is the theoretical yield of this reaction if 24.8 g CaCO3 is heated to yield 13.1 g CaO?

Handle like a typical stoichiometry problem: grams to grams

2) What is the percent yield?

Ignore the 13.1 g CaO to solve. This is the actual yield.

Chlorobenzene is used in the production of many different chemicals, such as aspirin, dyes and disinfectants. One method of preparing chlorobenzene is to react benzene C6H6, with chlorine according to the following equation:

1) When 36.8 g of C6H6 react with excess of Cl2, the actual yield of C6H5Cl is 38.8g?

What is the theoretical yield?

Handle like a typical stoichiometry problem: grams to grams

2) What is the percent yield?

C6H6 + Cl2 C6H5Cl + HCl

Standard Molar VolumeEqual volumes of all gases at the same temperature and pressure contain the same number of molecules.

- Amedeo Avogadro

At STP (Standard Temperature and Pressure):

1 mole of a gas occupies 22.4 liters of volume

Gas Stoichiometry #1If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.

3 H2(g) + N2(g) 2NH3(g)

3 moles H2 + 1 mole N2 2 moles NH3

3 liters H2 + 1 liter N2 2 liters NH3

Gas Stoichiometry #2How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?

3 H2(g) + N2(g) 2NH3(g)

12 L H2

L H2

= L NH3 L NH3

3

28.0

Gas Stoichiometry #3How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

= L O2

50.0 g KClO3 1 mol KClO3

122.55 g KClO3

3 mol O2

2 mol KClO3

22.4 L O2

1 mol O2

13.7