table of contents chapter 9 stoichiometry section 1 calculating quantities in reactions section 2...
TRANSCRIPT
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.
Section 1 Calculating Quantities in Reactions
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
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.
Problems Involving Mass
Example
What mass of NH3 can be made from 1221 g H2 and excess N2?
N2 + 3H2 2NH3
Section 1 Calculating Quantities in ReactionsChapter 9
Problems Involving Mass
Example
Section 1 Calculating Quantities in Reactions
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
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)
Section 1 Calculating Quantities in ReactionsChapter 9
Problems Involving Volume, continued
example
Section 1 Calculating Quantities in Reactions
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.
Chapter 9Section 1 Calculating Quantities in Reactions
Problems Involving Particles
Example
How many grams of C5H8 form from 1.89 × 1024 molecules C5H12?
C5H12(l) C5H8(l) + 2H2(g)
Section 1 Calculating Quantities in ReactionsChapter 9