Chapter 12Chapter 12

StoichiometryStoichiometry

Anything in black letters = write it in your notes (‘knowts’)

12.1 – The Arithmetic of Equations12.1 – The Arithmetic of Equations

12.2 – Chemical Calculations12.2 – Chemical Calculations

12.3 – Limiting Reagent & Percent Yield12.3 – Limiting Reagent & Percent Yield

12.1 – The Arithmetic of Equations12.1 – The Arithmetic of Equations

Quantitative study of chemical rxns.

In this chapter we will be asking ‘how much?’

Stoichiometry -

The coefficients of a balanced chemical equation represent the number of moles

that are reacting or produced.

2H2 + O2 2H2O

N2 + 3H2 2NH3

1 ( ) + 3 ( ) 2 ( )6.02 1023

molecules N2

6.02 1023

molecules H2

6.02 1023

molecules NH3

1 N2 molecule + 3 H2 molecules 2 NH3 molecules

It is even more practical to talk about moles…

It is not practical to talk about single molecules; instead use a larger number of molecules…

1 mol of N2 + 3 mol of H2 2 mol of NH3

The bike example…

For simplicity, say a bike requires only two things: 1 frame and 2 wheels.

2wheels + 1frame 1bike

What are the coefficients here? What do they tell you?

How many frames would be needed to ‘react’ completely with 20 wheels?

How many bikes could be produced from 4 wheels and 560 frames? What is the limiting reactant here?

+2wheels + 1frame 1bike

How many bikes would be produced from 23.7 kg of wheels and 80.1 kg of frames?

What is needed in order to solve the above question?

Always convert to a number of things (mol) first!

+2wheels + 1frame 1bike

ASSIGNMENT:

Chapter 12 Worksheet #1

12.2 – Chemical Calculations12.2 – Chemical Calculations

A mole ratio is a conversion factor that comes from the coefficients of a balanced chemical equation.

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

2 mol NH3

1 mol N2

1 mol N2

3 mol H2

3 mol H2

2 mol NH3

Write the three mole ratios that can be written from this balanced equation…

2 mol NH3

1 mol N2

1 mol N2

3 mol H2

3 mol H2

2 mol NH3

These are equivalent ratios, just upside down…

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

Mole ratios are used to convert from a given number of moles of a reactant or product to

moles of a different reactant or product.

Example 1 : How many moles of NH3 are produced when 0.60 mol of N2 reacts with excess H2?

0.60 mol N2 2 mol NH3

1 mol N2

= 1.2 mol NH3

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

Example 2:

Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.

2 mol NH3

3 mol H25.40 g H2

1 mol H2

2.0 g H2

17.0 g NH3

1 mol NH3

Given quantity

Change given unit to

moles

Mole ratio Change moles to grams

= 31 g NH3

Phosphorus burns in air to produce a phosphorus oxide in the following reaction:

4P(s) + 5O2(g) P4O10(s)

What mass of phosphorus will be needed to produce 3.25 mol of P4O10?

3.25 mol P4O10 31.0 g P

1 mol P= 403 g P

1 mol P4O10

4 mol P

How do you get better at something?

Did you tie your shoe correctly the very first time?Did you recite the alphabet correctly the first time?

Try the greek alphabet

How to Solve Stoichiometric Problems - Streamlined

1. Convert given # into moles, if it isn’t already

2. Multiply by the mole ratio conversion factor

3. Convert from moles of substance into desired unit if necessary.

What mass of frames would be needed to ‘react’ completely with 3060 g wheels

+2wheels + 1frame 1bike

45 g/wheel 27 g/frame 117 g/bike

3060 g wheels 1 frame

2 wheel

= 918 g frames

45 g wheels

1 wheel 27 g frame

1 frame

ASSIGNMENT:

Chapter 12 #11-16; 21-25 (p.391-398)

ASSIGNMENT:

Chapter 12 #45-50(p.411)

12.3 – Limiting Reagent & Percent Yield12.3 – Limiting Reagent & Percent Yield

The substance that is completely used up in a chemical rxn is called the limiting reagent.

A reagent is also known as a reactant

The substance that is NOT completely used up (and partially remains) is the excess reactant.

Example: Copper reacts with sulfur to form copper(I) sulfide. What is the limiting reagent when 80.0 grams of Cu react with 25.0 g S?

2Cu + S Cu2S

80.0 g Cu 63.5 g Cu1 mol Cu

1. Calculate the amount of one reactant required to react with the other.

2. Compare the given amount to the required amount.

2 mol Cu1 mol S

1 mol S32.06 g S

= 21.2 g Srequired amount.

0.630 mol S is needed to react with 1.26 mol Cu, there was 0.779 mol S given in the question.

Sulfur is in excess, so Cu is the limiting reagent

3. Compare the given amount to the required amount.

It doesn’t matter which reactant you use. If you used the actual amount of moles of S to find the amount of copper needed, then you would still

identify copper as the limiting reagent.

You Try It!

2Fe + O2 + 2H2O 2Fe(OH)2

If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?

2. Compare the given amount to the required amount.

1. Calculate the amount of one reactant required to react with the other.

The theoretical yield is the calculated amount of product that could be formed from given amounts of reactants; it is a the maximum amount.

The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield; it is usually lower than the theoretical yield

percent yield =actual yield

theoretical yield 100%

CS2 + 3Cl2 CCl4 + S2Cl2

You Try It!

What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2?

3.12 105 g CS2 76.142 g CS2

1 mol CS2 1 mol CCl41 mol CS2

153.81 g CCl41 mol CCl4

= 6.30 105 g CCl4 = 630 kg CCl4

Percent yield = 100% = 97.9%

617 kg CCl4630 kg CCl4

ASSIGNMENT:

Chapter 12 #26-38 (p.402 - 408)

1 ream of paper = 500 sheets

1 ream of copy paper = 5 lbs

1 ream of card-stock paper = 12 lbs

(1) 10 lbs of copy paper = ________ reams

(2) 1.31 lbs of copy paper = _______ reams

a paper analogy…

1 ream of paper = 500 sheets

1 ream of copy paper = 5 lbs

1 ream of card-stock paper = 12 lbs

(3) 0.23 reams card stock paper = _____ lbs

(4) 5.4 lbs of card stock paper = _______ sheets