Lesson 11Lesson 11 StoichiometryStoichiometry

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

11.1 – How Much??11.1 – How Much??

11.2 – Mole Ratios11.2 – Mole Ratios

11.3 – Excess and Limiting Reactants11.3 – Excess and Limiting Reactants

11.4 – Percent Yield11.4 – Percent Yield

11.5 – Molarity11.5 – Molarity

11.1 – How Much??11.1 – How Much??

Quantitative study of chemical rxns.

In this lesson 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

2H2 + O2 2H2O

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

molecules H2

6.02 1023

molecules O2

6.02 1023

molecules H3O

2 H2 molecules + 1 O2 molecule 2 H2O 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…

2 mol of H2 + 1 mol of O2 → 2 mol of H2O

The bike example…

For simplicity, say a bike requires only 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

Wheels Frames Number of bikes that could be made

How many wheels and/or frames leftover?

4 4

15

5 20

2 mol 2 mol

Complete each row in the chart…

Mg O2MgO

ProducedExcess Reactants

2 mol 1 mol

2 mol 2 mol

0.24 mol

0.036 mol

2.0 grams

Complete each row in the chart using the given amounts

2Mg + O2 2MgO

11.2 – Mole Ratios11.2 – Mole Ratios

A mole ratio comes from the coefficients of a balanced chemical equation.

Mole ratios are used to compare the amount of mol of one substance to another.

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)

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

How many moles of NH3 are produced when 1.0 mol of N2 reacts with excess H2?

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

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

= 31 g NH3

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

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

= 403 g P

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

= 918 g frames

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

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

11.3 – Excess and Limiting Reactants11.3 – Excess and Limiting Reactants

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

1.26 mol Cu

1. Convert given amounts into moles.

2. Multiply either amount by the mole ratio.

= 0.75 mol S24.0 g S

2 mol Cu

1 mol S

= 1.26 mol Cu

= 0.630 mol Cu

3. The smaller number of mol is the limiting reactant.

Cu is 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%

11.4 – Percent Yield11.4 – Percent Yield

CS2 + 3Cl2 CCl4 + S2Cl2

You Try It!

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

Practice Quiz Quiz

Practice

Learn from mistakes

Repeat

Check answer key

“Studying” is a myth

Best way to study is to re-do practice quiz

Use class time for work time and help

Some points can be recovered in 1:1 time with Tischer

Quarter Grades50% Tests/Quizzes35% Classwork/Homework15% Labs

Semester Grades45% 1st Qtr45% 2nd Qtr

10% Semester Test

Which of these matters the most??

Molarity – unit of solution concentration

mol of soluteMolarity =

Liters of solution

11.5 – Molarity11.5 – Molarity

1 Liter = 1000 mL

Calculate the molarity of these solutions.

a) 0.55 mol NaOH dissolved in 1.0 L solution

b) 4.0 grams of NaOH dissolved in 1.0 L solution

c) 4.0 grams of NaOH dissolved in 250 mL solution

How many moles of solute would be in the following solutions?

a) 1.00 Liter of 2.2 M HNO3

b) 25.0 mL of 0.225 M HCl

To make 1.0 L of a 1.0 M NaOH solution,

1.0 mol NaOH = 40.0 grams NaOH