Table of Contents Chapter 9 Stoichiometry Section 1 Calculating Quantities in Reactions Section 2 Limiting Reactants and Percentage Yield Section 3 Stoichiometry.

Table of Contents

Chapter 9

Stoichiometry

Section 1 Calculating Quantities in Reactions

Section 2 Limiting Reactants and Percentage Yield

Section 3 Stoichiometry and Cars

• Write the quantities of ingredients you would use to make a sandwich.

• Then determine how many sandwiches you could make from 24 slices of bread.

• Calculate how much of each other ingredient is needed.


Chapter 9

Objectives

• Use proportional reasoning to determine mole ratios from a balanced chemical equation.

• Explain why mole ratios are central to solving stoichiometry problems.

• Solve stoichiometry problems involving mass by using molar mass.

• Solve stoichiometry problems involving the volume of a substance by using density.

Chapter 9Section 1 Calculating Quantities in Reactions

Chapter 9: Stoichiometry

How much of a reactant is needed to produce a

given quantity of product

How much of product is formed from a given quantity of reactant

The branch of chemistry that deals with quantities of substances in chemical reactions is known as stoichiometry.

Stoichiometry is the proportional relationship between two or more substances during a chemical reaction.

Sec 1: Calculating quantities in reaction

4 different types of problems Mole mole Mass mass Volume Volume Particle Particle

Remember!

1 mol = 6.022 x particles Molar mass = mass (g) of 1 mol of

atoms (atomic mass) Volumes of Gases: at STP (standard

temperature and pressure) 1 mol of ANY gas = 22.4 L

2310

Interpreting Chemical equations

Balanced equation (given)

Particles: 2 atoms Na react with 1 molecule (2 atoms Cl) to form 2 molecules of NaCl.

Moles: 2 mol Na react with 1 mol (2 mol Cl) to form 2 mol of NaCl.

Mass: 2 mol Na x 22.99 g/mol = 45.98 g Na 2 mol Cl x 35.45 g/mol= 70.90 g 2 mol NaCl x (22.99+35.45)g/mol = 116.88 g

NaCl So, 45.98 g Na reacts with 70.90 g to form 116.88g

NaCl

2Cl

2Cl

2Cl

2Cl

)(2)()(2 2 sNaClgClsNa

Mole Ratios The only way to relate different chemicals. Must have a balanced equation.

2 mol NO reacts with 1 mol O to form 2 mol NO No matter how much you have of each, the same

Ratio will react. The following are Mole Ratios for the reaction above.

2 mol NO 1 mol O2 2 mol NO2 1 mol O2 2 mol NO2 2 mol NO

)(2)()(2 22 gNOgOgNO

Interpret this equation Hint: Balance first!

Particles: Moles: Masses:

)()()()( 22262 lOHgCOgOgHC

Mole Mole problems

Use mole ratios to solve problems.

How many mol of O2 will react with 7.22 mol NO?

)(2)()(2 22 gNOgOgNO

This process is called Stoichiometry– figuring out quantities of substances in chemical reaction

The steps of Stoichiometry Start with what you have and convert

to mole. Use the mole ratio from the balanced

chemical equation to find the moles of the new substance you’re interested in.

Convert from moles to desired units.

Solving Mass-Mass Problems

Section 1 Calculating Quantities in ReactionsChapter 9

Problems Involving Mass

Example

What mass of NH3 can be made from 1221 g H2 and excess N2?

N2 + 3H2 2NH3


Problems Involving Mass

Example


3 32

3 22 2 3

2 mol NH 17.04 g NH1mol H? g NH 1221 g H

2.02 g H 3 mol H 1mol NH

36867 g NH

Chapter 9

Solving Volume-Volume Problems


Problems Involving Volume

Example

What volume of H3PO4 forms when 56 mL POCl3 completely react? (density of POCl3 = 1.67 g/mL; density of H3PO4 = 1.83 g/mL)

POCl3(l) + 3H2O(l) H3PO4(l) + 3HCl(g)


Problems Involving Volume, continued

example


33 4 3

3

3 3 4 3 4

3 3 3 4

3 4

3 4

1.67 g POCl? mL H PO 56 mL POCl

1mL POCl

1mol POCl 1mol H PO 98.00 g H PO

153.32 g POCl 1mol POCl 1mol H PO

1mL H PO

1.83 g H PO

3 433 mL H PO

Chapter 9

Problems Involving Particles,

For Number of Particles, Use Avogadro’s Number• You can use Avogadro’s number,

6.022 1023 particles/ mol, in stoichiometry problems.

• If you are given particles and asked to find particles, Avogadro’s number cancels out.

• For this kind of calculation you use only the coefficients from the balanced equation.


Solving Particle Problems


Problems Involving Particles

Example

How many grams of C5H8 form from 1.89 × 1024 molecules C5H12?

C5H12(l) C5H8(l) + 2H2(g)


Problems Involving Particles, continued

Example


245 8 5 12

5 12 5 8 5 823

5 12 5 12 5 8

? g C H 1.89 10 molecules C H

1mol C H 1mol C H 68.13 g C H

6.022 10 molecules C H 1mol C H 1mol C H

5 8214 g C H

Table of Contents Chapter 9 Stoichiometry Section 1 Calculating Quantities in Reactions Section 2 Limiting Reactants and Percentage Yield Section 3 Stoichiometry.

Documents

mol of nacl

mol o2

mol of o2

mol no2

mol nointerpret

mol na x

mol nacl x

mass g