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Stoichiometry. Chapter 9. Stoichiometry. In this chapter, we will revisit the mole concept and use it to relate quantities of reactants & products. Stoichiometry. What is stoichiometry? Outside world connections – cooking, manufacturing, etc. Interpreting balanced chemical equations. - PowerPoint PPT Presentation
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Stoichiometry
• In this chapter, we will revisit the mole concept and use it to relate quantities of reactants & products.
Stoichiometry
What is stoichiometry?
• Outside world connections – cooking, manufacturing, etc.
• Interpreting balanced chemical equations
StoichiometryWhat is stoichiometry?• Chemical equation consists of reactants &
products, coefficients are there to give # of particles involved (moles, molecules, formula units, atoms, etc.)– Will later use coefficients from balanced
equations to find actual amounts of substances (mass, volume, etc.)
Stoichiometry Example
French Toast
• 3 eggs, slightly beaten
• 1 c milk
• ½ tsp salt
• 6 slices of bread
• 2 Tbs sugar
Stoichiometry ExampleFrench Toast• The recipe makes 6 slices of French
Toast. How would you make 12 slices?
• What is the ratio of eggs to bread? Did it change when the recipe was increased to make more servings?
• What would happen if you tried to make the recipe with only 1 egg?
Stoichiometry Example
Following a recipe is a lot like performing a chemical reaction. You have to have the right reactants and enough of each one or things will not turn out as planned.
Stoichiometry…
Ratios are very important in balanced chemical equations.
• What is the ratio of H2 to O2 in the following equation? 2 H2 + O2 2 H2O
• Would the ratio change if more water was produced? Why?
Stoich Problems
How do I use stoichiometry?
• Follow the process for solving problems: qty of given moles of given moles of unknown qty of unknown
Beginning Stoich
Mole-mole problems
• Converting from moles of one substance to moles of another substance requires a mole ratio from the balanced chemical equation
Beginning Stoich
Mole-mole problemsEx 1:• A pink paint is produced by using 3 L
of white & 1 L of red–The paint ratio is ______–If we only want 2 L of pink paint
what do we need?•Mix ___ L white & ___ L red
Beginning Stoich
Mole-mole problems
Ex 2:
• NH4NO3 N2O + 2 H2O
• Suppose we wanted to know the number of moles of product created by 2.25 moles ammonium nitrate
Beginning Stoich
Mole-mole problems
Ex 3:
• 2 HCl + Zn ZnCl2 + H2
• How many moles HCl needed to react with 5.70 moles Zn?
Beginning Stoich
Mole-mole problems
Ex 4:
• N2O5 + H2O 2 HNO3
• How many moles HNO3 produced from 0.51 moles N2O5?
Beginning Stoich
Mole-mole problems
Ex 5:
• Pb + 2 HCl PbCl2 + H2
• How many moles HCl needed to react with 0.36 moles Pb?
Beginning Stoich
Verifying the law of conservation of matter
Balanced eqs “obey” the law:• 2 H2 + O2 2 H2O
• 2 moles H2 (@ 2.016 g/mole) = g• 1 mole O2 (@ 32.00 g/mole) = g• 2 moles water (@ 18.016 g/mole) = g
Stoich Problems
4 types of problems:1) Mass-mass – given mass of one
substance & asked to find mass of another
• Process: (g of given) * (1 mole/molar mass of given) * (ratio of wanted to given) * (molar mass of wanted/1 mole)
Stoich Problems
4 types of problems:
1) Mass-mass
Example 1: How many grams of AgCl can be produced when 17.0 g AgNO3 reacts with NaCl?
Stoich Problems
4 types of problems:
1) Mass-mass
Example 2: How many grams of Cu2S can be produced from 9.90 g of CuCl reacting with H2S?
Stoich Problems
4 types of problems:2) Mass-volume - given mass of one
substance & asked to find volume of another
• Process: (g of given) * (1 mole/molar mass of given) * (ratio of wanted to given) * (molar volume/1 mole)
Stoich Problems
4 types of problems:
2) Mass-volume
Example 1: What volume of H2 can be produced (at STP) when 6.54 g Zn reacts with HCl?
Stoich Problems
4 types of problems:
2) Mass-volume
Example 2: What volume of ammonia can be produced from 14.01 g N2 reacting with H2?
Stoich Problems
4 types of problems:3) Volume-mass - given volume of
one substance & asked to find mass of another
• Process: (volume of given) * (1 mole/molar volume of given) * (ratio of wanted to given) * (mass/1 mole)
Stoich Problems
4 types of problems:
3) Volume-mass -
Example 1: How many grams of NaCl can be produced by reacting 112 mL Cl2 (@ STP) w/ Na?
Stoich Problems
4 types of problems:
3) Volume-mass -
Example 2: Bromine reacts with 5600 mL hydrogen to give what mass of HBr at STP?
Stoich Problems
4 types of problems:4) Volume-volume - given volume of
one substance & asked to find volume of another
• Process: (volume of given) * (1 mole/molar volume of given) * (ratio of wanted to given) * (molar volume/1 mole)
Stoich Problems
4 types of problems:
4) Volume-volume -
Example 1: How many L of O2 are needed to burn 1.00 L of methane (@ STP)?
Stoich Problems
4 types of problems:
4) Volume-volume -
Example 2: What volume of Br2 is produced if 75.2 L of Cl2 react with HBr?
Limiting Reactant
Limiting reactant – substance that determines how much product may be formed (or how much reactant was available)
Limiting ReactantLimiting reactant –• Identifying –
1) Determine possible amount of desired substance (need to do 2 dimensional analysis problems)– The limiting reactant is the
substance that results in the least amount of desired substance
Limiting Reactant
Limiting reactant Problems
Ex. 1 2 Fe + 3 S Fe2S3
• Given 111.7 g Fe & 160.35 g S, how many grams of product can be formed?
Limiting Reactant
Limiting reactant Problems
Ex. 2 Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag
• Given 3.5 g copper & 6.0 grams silver nitrate, how much silver can be formed?
Limiting Reactant
Limiting reactant Problems
Ex. 3 Zn + S ZnS
• Given 3.5 g each reactant, how much product formed?
Percent Yield
• how much product is formed compared with expected results
% yield = (actual amount/expected amount) x 100
(units are in terms of %)
Percent Yield
Ex. 1) You combined 4.3 liters of Cl2 with some Na. You were able to recover 15.9 g of NaCl. What was the percent yield?
• 2 Na + Cl2 2 NaCl
Percent Yield
Ex. 2) 5.00 grams of Cu mixed with excess AgNO3. This gave 15.2 g of Ag. What was the percent yield?
• Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag