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Cookies and Chemistry…Cookies and Chemistry…Huh!?!?Huh!?!?
Just like chocolate chip Just like chocolate chip cookies have recipes, cookies have recipes, chemists have recipes as chemists have recipes as wellwell
Instead of calling them Instead of calling them recipes, we call them recipes, we call them reaction equationsreaction equations
Furthermore, instead of Furthermore, instead of using cups and teaspoons, using cups and teaspoons, we use moleswe use moles
Lastly, instead of eggs, Lastly, instead of eggs, butter, sugar, etc. we use butter, sugar, etc. we use chemical compounds as chemical compounds as ingredientsingredients
Chemistry RecipesChemistry Recipes Looking at a reaction tells us how Looking at a reaction tells us how
much of something you need to react much of something you need to react with something else to get a product with something else to get a product (like the cookie recipe)(like the cookie recipe)
Be sure you have a balanced reaction Be sure you have a balanced reaction before you start!before you start!
Example: 2 Na + ClExample: 2 Na + Cl2 2 2 NaCl 2 NaCl This reaction tells us that by mixing 2 moles This reaction tells us that by mixing 2 moles
of sodium with 1 mole of chlorine we will get of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride2 moles of sodium chloride
What if we wanted 4 moles of NaCl? 10 What if we wanted 4 moles of NaCl? 10 moles? moles? 50 moles?50 moles?
2H2H2 2 + O+ O22 2H2H22O O Two molecules of hydrogen and one Two molecules of hydrogen and one
molecule of oxygen form two molecules of molecule of oxygen form two molecules of water.water.
2 Al2 Al22OO33 AlAl ++ 3O3O22
2 formula units Al2O3 form 4 atoms Al
and 3 molecules O2
Now try this: 2Na + 2H2O 2NaOH + H2
PracticePractice Write the balanced reaction for hydrogen gas Write the balanced reaction for hydrogen gas
reacting with oxygen gas.reacting with oxygen gas. 2 H2 H22 + O + O22 2 H 2 H22OO
How many moles of reactants are needed?How many moles of reactants are needed? What if we wanted 4 moles of water?What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much What if we had 3 moles of oxygen, how much
hydrogen would we need to react and how much hydrogen would we need to react and how much water would we get?water would we get?
What if we had 50 moles of hydrogen, how much What if we had 50 moles of hydrogen, how much oxygen would we need and how much water oxygen would we need and how much water produced? produced?
Mole RatiosMole Ratios
These mole ratios can be used to These mole ratios can be used to calculate the moles of one chemical calculate the moles of one chemical from the given amount of a different from the given amount of a different chemical chemical
Example: How many moles of chlorine Example: How many moles of chlorine is needed to react with 5 moles of is needed to react with 5 moles of sodium (without any sodium left over)?sodium (without any sodium left over)?
2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl5 moles Na 1 mol Cl2
2 mol Na= 2.5 moles Cl2
Mole-Mole ConversionsMole-Mole Conversions How many moles of sodium chloride How many moles of sodium chloride
will be produced if you react 2.6 will be produced if you react 2.6 moles of chlorine gas with an excess moles of chlorine gas with an excess (more than you need) of sodium (more than you need) of sodium metal?metal?
Mole-Mass ConversionsMole-Mass Conversions
Most of the time in chemistry, the amounts Most of the time in chemistry, the amounts are given in grams instead of molesare given in grams instead of moles
We still go through moles and use the mole We still go through moles and use the mole ratio, but now we also use molar mass to get ratio, but now we also use molar mass to get to gramsto grams
Example: How many grams of chlorine are Example: How many grams of chlorine are required to react completely with 5.00 moles of required to react completely with 5.00 moles of sodium to produce sodium chloride?sodium to produce sodium chloride?
2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl
5.00 moles Na 1 mol Cl2 70.90g Cl2
2 mol Na 1 mol Cl2
= 177g Cl2
PracticePractice
Calculate the mass in grams of Calculate the mass in grams of Iodine required to react completely Iodine required to react completely with 0.50 moles of aluminum.with 0.50 moles of aluminum.
Mass-MoleMass-Mole We can also start with mass and convert We can also start with mass and convert
to moles of product or another reactantto moles of product or another reactant We use molar mass and the mole ratio to We use molar mass and the mole ratio to
get to moles of the compound of interestget to moles of the compound of interest Calculate the number of moles of ethane Calculate the number of moles of ethane
(C(C22HH66) needed to produce 10.0 g of water) needed to produce 10.0 g of water 2 C2 C22HH66 + 7 O + 7 O22 4 CO 4 CO22 + 6 H + 6 H220 0
10.0 g H2O 1 mol H2O 2 mol C2H6
18.0 g H2O 6 mol H20
= 0.185 mol C2H6
PracticePractice Calculate how many moles of oxygen Calculate how many moles of oxygen
are required to make 10.0 g of are required to make 10.0 g of aluminum oxidealuminum oxide
Mass-Mass ConversionsMass-Mass Conversions Most often we are given a starting Most often we are given a starting
mass and want to find out the mass mass and want to find out the mass of a product we will get (called of a product we will get (called theoretical yield) or how much of theoretical yield) or how much of another reactant we need to another reactant we need to completely react with it (no leftover completely react with it (no leftover ingredients!)ingredients!)
Now we must go from grams to Now we must go from grams to moles, mole ratio, and back to grams moles, mole ratio, and back to grams of compound we are interested inof compound we are interested in
Mass-Mass ConversionMass-Mass Conversion
Ex. Calculate how many grams of Ex. Calculate how many grams of ammonia are produced when you ammonia are produced when you react 2.00g of nitrogen with excess react 2.00g of nitrogen with excess hydrogen.hydrogen.
NN2 2 + 3 H+ 3 H2 2 2 NH 2 NH332.00g N2 1 mol N2 2 mol NH3 17.06g NH3
28.02g N2 1 mol N2 1 mol NH3
= 2.4 g NH3
PracticePractice
How many grams of calcium nitride How many grams of calcium nitride are produced when 2.00 g of calcium are produced when 2.00 g of calcium reacts with an excess of nitrogen?reacts with an excess of nitrogen?
Mole to Mole conversionsMole to Mole conversions 2 Al2 Al22OO33 AlAl ++ 3O3O22
each time we use 2 moles of Aleach time we use 2 moles of Al22OO33 we will also we will also make 3 moles of Omake 3 moles of O22
2 moles Al2O3
3 mole O2
or2 moles Al2O3
3 mole O2
These are the two possible conversion factors
Mole to Mole conversionsMole to Mole conversions How many moles of OHow many moles of O22 are are
produced when 3.34 moles of produced when 3.34 moles of AlAl22OO33 decompose? decompose?
2 Al2 Al22OO33 AlAl ++ 3O3O22
3.34 mol Al2O3 2 mol Al2O3
3 mol O2 = 5.01 mol O2
Practice:Practice: 2C2C22HH22 + 5 O + 5 O22 4CO 4CO22 + 2 H + 2 H22OO
• If 3.84 moles of CIf 3.84 moles of C22HH22 are burned, are burned, how many moles of Ohow many moles of O22 are are needed?needed?
(9.6 mol)
•How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol)
•If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol)
Mass-Mass Problem:Mass-Mass Problem:
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al + 3 O2 2Al2O3
=6.50 g Al
? g Al2O3
1 mol Al
26.98 g Al 4 mol Al
2 mol Al2O3
1 mol Al2O3
101.96 g Al2O3
(6.50 x 2 x 101.96) ÷ (26.98 x 4) = 12.3 g Al2O3
Another example:Another example: If 10.1 g of Fe are added to a If 10.1 g of Fe are added to a
solution of Copper (II) Sulfate, solution of Copper (II) Sulfate, how much solid copper would how much solid copper would form?form?
2Fe + 3CuSO2Fe + 3CuSO44 Fe Fe22(SO(SO44))33 + + 3Cu3Cu Answer = 17.2 g Cu
Volume-Volume Volume-Volume Calculations:Calculations:
How many liters of CHHow many liters of CH4 4 at STP are required to at STP are required to
completely react with 17.5 L of Ocompletely react with 17.5 L of O2 2 ??
CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO
17.5 L O2 22.4 L O2 1 mol O2
2 mol O2
1 mol CH4
1 mol CH4 22.4 L CH4
= 8.75 L CH4
22.4 L O2 1 mol O2
1 mol CH4 22.4 L CH4
Notice anything concerning these two steps?
50.0 mL
6.0 M
L
mol 6.0
? g
50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?
H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)
Solution Stoichiometry
=
Our Goal
= g NaHCO3
H2SO4
50.0 mL
1000mL
SOH mol 6.0
42SOH
42
1 molH2SO4
NaHCO3
2 molNaHCO3
84.0 gmolNaHCO3
50.4
When N2O5 is heated, it decomposes:
2N2O5(g) 4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
= moles NO2
4.3 mol N2O5
52
2
ON mol2
NO mol48.6
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
= mole O2
4.3 mol N2O5
52
2
ON 2mol
O mol12.2
2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol
2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol
Mole – Mole Conversions
Units match
When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
= moles N2O5
210 g NO2
2
52
NO mol4
ON mol22.28
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
= grams N2O5
75.0 g O2
2
52
O 1mol
ON mol2506
2
2
NO g0.46
NO mol
2
2
O g 32.0
O mol
52
52
ON mol
ON g108
gram ↔ mole and gram ↔ gram conversions
2N2O5(g) 4NO2(g) + O2(g)210g? moles
2N2O5(g) 4NO2(g) + O2(g)75.0 g? grams
Units match
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams
Let’s work the problem.
= g AlCl3
3.45 g Al
Alg 27.0
Almol
We must always convert to moles.
Now use the molar ratio.
Almol 2
AlClmol 2 3
Now use the molar mass to convert to grams.
3
3
AlClmol
AlClg 133.317.0
Units match
gram to gram conversions
What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) + Ba(OH)2(aq) 2H2O(l) + BaCl2
0.40 M 47.1 mL0.75 M? mL
= mL HCl
Ba(OH)2
47.1 mL
2
2
Ba(OH)
Ba(OH)
mL 1000
0.75mol
1 mol Ba(OH)2
HCl2 mol
0.40 mol HCl
HCl1000 mL
176
Units match
Solution Stoichiometry
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?
____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1
23.28 mL
0.135 mol L
25.00 mL
? mol L
= mol Ba(OH)2
L Ba(OH)2
25.00 x 10-3 L Ba(OH)2
Units Already Match on Bottom!
HClmL 23.28
HCl
HCl
mL 1000
mol 0.135
HCl
Ba(OH)
mol 2
mol l2 0.0629
Units match on top!
Limiting Reactant: CookiesLimiting Reactant: Cookies1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chipsMakes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?
Limiting ReactantLimiting Reactant
Most of the time in chemistry we have Most of the time in chemistry we have more of one reactant than we need to more of one reactant than we need to completely use up other reactant.completely use up other reactant.
That reactant is said to be in That reactant is said to be in excessexcess (there is too much).(there is too much).
The other reactant limits how much The other reactant limits how much product we get. Once it runs out, the product we get. Once it runs out, the reaction s. This is called the reaction s. This is called the limiting reactantlimiting reactant..
Limiting ReactantLimiting Reactant To find the correct answer, we have to try To find the correct answer, we have to try
allall of the reactants. We have to calculate of the reactants. We have to calculate how much of how much of aa product we can get from product we can get from eacheach of the reactants to determine which of the reactants to determine which reactant is the limiting one.reactant is the limiting one.
The The lowerlower amount of amount of aa product is the product is the correct answer.correct answer.
The reactant that makes the least amount The reactant that makes the least amount of product is the of product is the limiting reactantlimiting reactant. Once . Once you determine the limiting reactant, you you determine the limiting reactant, you should ALWAYS start with it!should ALWAYS start with it!
Be sure to pick Be sure to pick aa product! You can’t product! You can’t compare to see which is greater and which compare to see which is greater and which is lower unless the product is the same!is lower unless the product is the same!
Limiting Reactant: ExampleLimiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of 10.0g of aluminum reacts with 35.0 grams of
chlorine gas to produce aluminum chloride. chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, Which reactant is limiting, which is in excess, and how much product is produced?and how much product is produced?
2 Al + 3 Cl2 Al + 3 Cl22 2 AlCl 2 AlCl33 Start with Al:Start with Al:
Now ClNow Cl22::
10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3
27.0 g Al 2 mol Al 1 mol AlCl3
= 49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3
71.0 g Cl2 3 mol Cl2 1 mol AlCl3
= 43.9g AlCl3
LimitingLimitingReactantReactant
LR Example ContinuedLR Example Continued We get We get 49.4g49.4g of aluminum chloride from the of aluminum chloride from the
given amount of aluminum, but only given amount of aluminum, but only 43.9g43.9g of aluminum chloride from the given of aluminum chloride from the given amount of chlorine. Therefore, chlorine is amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a chlorine is used up, the reaction comes to a complete .complete .
Limiting Reactant PracticeLimiting Reactant Practice
15.0 g of potassium reacts with 15.0 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant g of iodine. Calculate which reactant is limiting and how much product is is limiting and how much product is made.made.
If 10.6 g of copper reacts with 3.83 g If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product sulfur, how many grams of product (copper (I) sulfide) will be formed?(copper (I) sulfide) will be formed?
2Cu + S 2Cu + S Cu Cu22SS
10.6 g Cu 63.55g Cu 1 mol Cu
2 mol Cu 1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 13.3 g Cu2S
3.83 g S 32.06g S 1 mol S
1 mol S 1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 19.0 g Cu2S
= 13.3 g Cu2S
Cu is Limiting Reagent
Finding the Amount of Finding the Amount of ExcessExcess
By calculating the amount of the By calculating the amount of the excess reactant needed to excess reactant needed to completely react with the limiting completely react with the limiting reactant, we can subtract that reactant, we can subtract that amount from the given amount to amount from the given amount to find the amount of excess.find the amount of excess.
Can we find the amount of excess Can we find the amount of excess potassium in the previous problem?potassium in the previous problem?
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.
4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)
0.15 mol 0.10 mol ? moles
Based on:KO2 = mol O2
0.15 mol KO2
2
2
KO 4mol
O mol30.1125
Hide
Based on: H2O
= mol O20.10 mol H2O
OH 2mol
O mol3
2
2 0.150
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)
0.15 mol 0.10 mol ? moles
Based on:KO2 = mol O2
0.15 mol KO2
2
2
KO 4mol
O mol30.1125
Based on: H2O
= mol O20.10 mol H2O
OH 2mol
O mol3
2
2 0.150
What is the theoretical yield? Hint: Which is the smallest
amount? The is based upon the limiting reactant?
It was limited by theamount of KO2.
H2O = excess (XS) reactant!
Finding Excess PracticeFinding Excess Practice 15.0 g of potassium reacts with 15.0 g of 15.0 g of potassium reacts with 15.0 g of
iodine. iodine. 2 K + I2 K + I22 2 KI 2 KI
We found that Iodine is the limiting We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide reactant, and 19.6 g of potassium iodide are produced.are produced.15.0 g I2 1 mol I2 2 mol K 39.1 g K
254 g I2 1 mol I2 1 mol K= 4.62 g K USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!
Limiting Reactant: RecapLimiting Reactant: Recap
1.1. You can recognize a limiting reactant problem You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.because there is MORE THAN ONE GIVEN AMOUNT.
2.2. Convert ALL of the reactants to the SAME product Convert ALL of the reactants to the SAME product (pick any product you choose.)(pick any product you choose.)
3.3. The lowest answer is the correct answer.The lowest answer is the correct answer.4.4. The reactant that gave you the lowest answer is the The reactant that gave you the lowest answer is the
LIMITING REACTANT.LIMITING REACTANT.5.5. The other reactant(s) are in EXCESS.The other reactant(s) are in EXCESS.6.6. To find the amount of excess, subtract the amount To find the amount of excess, subtract the amount
used from the given amount.used from the given amount.7.7. If you have to find more than one product, be sure If you have to find more than one product, be sure
to start with the limiting reactant. You don’t have to start with the limiting reactant. You don’t have to determine which is the LR over and over again!to determine which is the LR over and over again!