III. Stoichiometry Stoy – kee – ahm –eh  tree. Chapter 12. Sections Click the section to jump to the slides. Mole Ratios MoletoMole Calculations MoletoMass Calculations Particle Calculations Molar Volume Calculations Limiting Reactants Percent Yields. Things you should remember.
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Chapter 12
2
__H2SO4(aq) + __NaHCO3(s)
__Na2SO4(aq) + __H2O(l) + __CO2(g)
2
2
What’s the relationship between NaHCO3 and Na2SO4?
2 mole NaHCO3 = 1 mol Na2SO4
(it’s just the coefficients!!)
What’s the relationship between H2SO4 and CO2?
1 mol H2SO4 = 2 mol CO2
(it’s just the coefficients!!)
or
or
2 mol/mc NaHCO3
or
1 mole/mc H2SO4
or
or
2 mol/mc CO2
or
1 mole/mc H2SO4
After writing a balanced equation, write down 3 possible relationships. (hint: just look at the coefficients.)
Fe + O2 Fe2O3
Skeletal equation:
4Fe + 3O2 2Fe2O3
Balanced equation:
4 mol Fe = 3 mol O2
Relationships:
4 mol Fe = 2 mol Fe2O3
3 mol O2 = 2 mol Fe2O3
1. Write three mole ratios (relationships) from the reaction below:
Al2S3 + H2O Al(OH)3 + H2S
6
2
3
1 mol Al2S3 = 6 mol H2O
1 mol Al2S3 = 2 mol Al(OH)3
1 mol Al2S3 = 3 mol H2S
You should have 3
6 mol H2O = 2 mol Al(OH)3
6 mol H2O = 3 mol H2S
2 mol Al(OH)3 = 3 mol H2S
1. Write three mole ratios (relationships) from the reaction below:
Al2S3 + H2O Al(OH)3 + H2S
6
2
3
1 mol Al2S3 = 6 mol H2O
1 mol Al2S3 = 2 mol Al(OH)3
1 mol Al2S3 = 3 mol H2S
6 mol H2O = 2 mol Al(OH)3
6 mol H2O = 3 mol H2S
2.Aluminum is produced by decomposing aluminum oxide into aluminum and oxygen.
a. Write a balanced equation.
b. Write all the molar ratios that can be derived from this equation.
2Al2O3 4Al + 3O2
2 mol Al2O3 = 4 mol Al
2 mol Al2O3 = 3 mol O2
4 mol Al = 3 mol O2
1. Write a balanced reaction.
2. Determine your given & want
3. Determine your relationships. If you see…
Now you are ready to solve IT’S GAME TIME.
N2 (g) + 3H2 (g) 2NH3 (g)
Haber Process: an industrial process for producing ammonia from nitrogen and hydrogen by combining them under high pressure in the present of an iron catalyst
source: worldnet.princeton.edu
Haber Process:
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
No grams, No molar mass!!
Given :
Want :
Relationships:
4.00 mol H2
No fun, mc, No 6.02 x 1023!!
? mol NH3
2 mol NH3 = 3 mol H2
4.00 mol H2
2 mol NH3
= mol NH3
2.67
x
3 mol H2
1
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
No grams, No molar mass!!
Given :
Want :
Relationships:
7.8 mol NH3
No fun, mc, No 6.02 x 1023!!
? mol N2
2 mol NH3 = 1 mol N2
7.8 mol NH3
1 mol N2
= mol N2
3.9
X
2 mol NH3
1
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
No grams, No molar mass!!
Given :
Want :
Relationships:
13 mol N2
No fun, mc, No 6.02 x 1023!!
? mol H2
1 mol N2 = 3 mol H2
13 mol N2
3 mol H2
= moles H2
39
x
1 mol N2
1
1. Write a balanced reaction.
2. Determine your given & want
3. Determine your relationships. If you see…
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
How do I know when to use mole ratios, molar mass or 6.02 x 1023?
Given :
Want :
Relationships:
4.00 mol H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 17.031g NH3
Look at the “given” and “want” for clues
No fun, mc, No 6.02 x 1023!!
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
4.00 mol H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 17.031g NH3
4.00 mol H2
2 mol NH3
17.031g NH3
x
x
3 mol H2
1 mol NH3
1
= g NH3
45.4
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? mol NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
Remember your Steps to Success!
No fun, mc, No 6.02 x 1023!!
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? mol NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
4.00 g H2
1 mol H2
2 mol NH3
x
x
2.016g H2
3 mol H2
1
= mol NH3
1.32
1. How many grams of nitrogen will react with 3.40 moles of hydrogen to produce ammonia?
2. How many grams of hydrogen are required to make 54.0 moles of ammonia in excess nitrogen?
3. How many moles of nitrogen react completely with 3.70 moles of hydrogen?
31.7 g N2
163 g H2
1.23 mol N2
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
Remember your Steps to Success!
1 mol NH3 = 17.031 g NH3
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
4.00 g H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = 17.031g NH3
17.031g NH3
1 mol H2
2 mol NH3
4.00 g H2
x
x
x
2.016g H2
3 mol H2
1 mol NH3
1
= g NH3
22.5
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
12.0 g H2
? g N2
1 mol H2 = 2.016g H2
Remember your Steps to Success!
1 mol N2 = 28.014 g N2
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
12.0 g H2
? g N2
1 mol N2 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol N2 = 28.014 g N2
28.014g N2
12.0 g H2
1 mol H2
1 mol N2
x
x
x
2.016g H2
1
3 mol H2
1 mol N2
= g N2
55.6
340 g NH3
8.22 x 104 g N2
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
2.09 x 1014 mc H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 17.031g NH3
1 mol H2 = 6.02 x 1023 mc H2
Ex 1: How many grams of ammonia are produced from 2.09 x 10 Hydrogen?14 mc of hydrogen gas?
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
2.09 x 1014 mc H2
? g NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 6.02 x 1023 mc H2
1 mol NH3 = 17.031g NH3
1 mol H2
2 mol NH3
2.09 x 1014 mc H2
17.031g NH3
x
x
x
3 mol H2
1
6.02 x 1023 mc H2
1 mol NH3
3.94 x 109
= g NH3
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
40.2g H2
? mc NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = 6.02 x 1023 mc NH3
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
40.2g H2
? mc NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = 6.02 x 1023 mc NH3
1 mol H2
40.2g H2
2 mol NH3
6.02 x 1023 mc NH3
x
x
x
2.016g H2
3 mol H2
1
1 mol NH3
8.00 x 1024
= mc NH3
7.01 x 1023 mc N2
4.18 x 1021 mc NH3
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
7.04 g H2
? mc N2
1 mol N2 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol N2 = 6.02 x 1023 mc N2
1 mol H2
7.04 g H2
1 mol N2
6.02 x 1023 mc N2
x
x
x
2.016g H2
3 mol H2
1
1 mol N2
7.01 x 1023
= mc N2
How many mc of ammonia will grams of hydrogen?2.09 x 1021 mc of N2 produce in excess hydrogen?
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
2.09 x 1013 mc N2
? mc NH3
2 mol NH3 = 1 mol N2
1 mol N2 = 6.02 x 1023 mc N2
1 mol NH3 = 6.02 x 1023 NH3
2.09 x 1013 mc N2
1 mol N2
2 mol NH3
6.02 x 1023 mc NH3
x
x
x
1
1 mol N2
1 mol NH3
6.02 x 1023 mc N2
4.18 x 1021
= mc NH3
1 mol __ = 22.4 L __
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
44.8 L H2
? mol N2
1 mol N2 = 3 mol H2
1 mol H2 = 22.4 L H2
Remember your Steps to Success!
No grams, No molar mass
No fun, mc, No 6.02 x 1023!!
Balanced Equation:
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
44.8 L H2
? mol N2
1 mol N2 = 3 mol H2
1 mol H2 = 22.4 L H2
44.8 L H2
1 mol H2
1 mol N2
= mol N2
0.667
x
x
22.4 L H2
3 mol H2
1
Given over 1
Mole ratio
1 mol = 22.4 L
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
5.00 mol H2
? L NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 22.4 L NH3
Remember your Steps to Success!
No grams, No molar mass
No fun, mc, No 6.02 x 1023!!
Ex 2: If 5.00 moles of H with 44.8 liters of hydrogen gas to produce ammonia gas?2 react with excess N2, how many liters of NH3 are produced?
N2 (g) + 3H2 (g) 2NH3 (g)
Given :
Want :
Relationships:
5.00 mol H2
? L NH3
2 mol NH3 = 3 mol H2
1 mol NH3 = 22.4 L NH3
2 mol NH3
5.00 mol H2
22.4 L NH3
= L NH3
74.7
x
x
3 mol H2
1 mol NH3
1
Given over 1
1 mol = 22.4 L
Mole ratio
“which reactant are you going to run out of first?”
“which reactant do you have extra of?”
4 is sometimes called the
+
4
13
Which is the limiting reactant
4 shells
1 car
= 4 cars can be made if 4 shells are available
1 shell
13 tires
1 car
= 3.25 cars can be made if 13 tires are available
4 tires
Less product can be produced, therefore tires must be the limiting reactant
Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant?
Available
Needed
1.21 mol Zn
2 mol HCl
= 2.42 mol HCl
1 mol Zn
2.64 mol HCl
1 mol Zn
= 1.32 mol Zn
2 mol HCl
Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant?
Available VS Needed
<
1.21 mol Zn 1.32 mol Zn
Limiting reactant
>
2.64 mol HCl 2.42 mol HCl
Excess reactant
1) Determine the balanced chemical equation
2) Convert the given/available amounts reactants to the number of moles of the other reactant(s)
3) Compare the available amounts to the needed amounts:
*If the available amount > needed amount, it is excess reactant
* If the available amount < needed amount, it is the limiting reactant
4) To calculate the amount of excess:
Available Amount – Needed Amount = Amount of Excess
5) To calculate the amount of product produced:
G: amount of available limiting reactant
W: amount of product
Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. Which is the limiting reactant when 0.75 mol of N2H4 is mixed with 0.50 mol of H2O2? How much excess reactant, in moles, remains unchanged? How much of each product, in moles, is formed?
Step 1: Write a balanced reaction 2HCl(aq)
Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam.
N2H4
+ H2O2
2
4
N2(g)
+ H2O(g)
N 2HCl(aq) 2H4
+ H2O2
2
4
N2(g)
+ H2O(g)
Available
Needed
Limiting Reactant
0.75 mol N2H4
2 mol H2O2
= 1.50 mol H2O2
1 mol N2H4
Excess
0.50 mol H2O2
1 mol N2H4
= 0.25 mol N2H4
2 mol H2O2
4. To calculate the amount of excess:
Available Amount – Needed Amount = Amount of Excess
Available 2HCl(aq)
Needed
0.75 mol N2H4
2 mol H2O2 =
1.50 mol H2O2
1 mol N2H4
0.50 mol H2O2
1 mol N2H4 =
Excess
0.25 mol N2H4
2 mol H2O2

= 0.50 mol N2H4
*Available – Needed = Remaining excess
5. To calculate the amount of product produced:
G: amount of available limiting reactant
W: amount of product
N 2HCl(aq) 2H4
+ H2O2
2
4
N2(g)
+ H2O(g)
G:W:R:
0.50 mol H2O2
How many moles of product are formed?
G: amount of available limiting reactant
mole N2
N2 is the first product
2 mol H2O2 = 1 mol N2
0.50 mol H2O2
1 mol N2
= 0.25 mol N2
2 mol H2O2
G:W:R:
0.50 mol H2O2
G: amount of available limiting reactant
H2O is the second product
mole H2O
2 mol H2O2 = 4 mol H2O
0.50 mol H2O2
4 mol H2O
= 1.0 mol H2O
2 mol H2O2
Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 2.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
Balanced Equation: 8 Zn + S8 8 ZnS
Available
Needed
2.0 mol Zn
1 mol S8
= 0.25 mol S8
8 mol Zn
1.0 mol S8
8 mol Zn
= 8.0 mol Zn
1 mol S8
Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
Available VS Needed
<
2.0 mol Zn 8.0 mol Zn
Limiting reactant
>
1.0 mol S8 0.25 mol S8
Excess reactant
Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
How much excess is present?
(Available Amount – Needed Amount = Amount of Excess)
1.0 mol S8 – 0.25 mol S8 = 0.75 mol S8 in excess
4. To calculate the amount of excess:
Available Amount – Needed Amount = Amount of Excess
5. To calculate the amount of product produced:
G: amount of available limiting reactant
W: amount of product
Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If two moles of zinc are heated with 1.00 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?
Balanced Equation: 8 Zn + S8 8 ZnS
How much product was formed?
G:W:R:
2 mol Zn
mol ZnS
8 mol Zn = 8 mol ZnS
2 mol Zn
8 mol ZnS
= 2 mol ZnS formed
8 mol Zn
Practice 2HCl(aq)
1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?
Balanced Equation: C + H2O H2 + CO
Available
Needed
2.4 mol C
1 mol H2O
= 2.4 mol H2O
1 mol C
3.1 mol H2O
1 mol C
= 3.1 mol C
1 mol H2O
Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?
Available VS Needed
<
2.4 mol C 3.1 mol C
Limiting reactant
>
3.1 mol H2O 2.4 mol H2O
Excess reactant
4. To calculate the amount of excess:
Available Amount – Needed Amount = Amount of Excess
5. To calculate the amount of product produced:
G: amount of available limiting reactant
W: amount of product
Balanced Equation: C + H produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, 2O H2 + CO
G:W:R:
2.4 mol C
Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?
G: amount of available limiting reactant
mole H2
H2 is the first product
1 mol C = 1 mol H2
2.4 mol C
1 mol H2
= 2.4 mol H2
1 mol C
G:W:R:
2.4 mol C
G: amount of available limiting reactant
CO is the second product
mole CO
1 mol C = 1 mol CO
2.4 mol C
1 mol CO
= 2.4 mol CO
1 mol C
2) Zn + Pb(NO3)2 Pb + Zn(NO3)2
3) Fe + 2HCl H2 + FeCl2
2NaOH + H2SO4 Na2SO4 + 2H2O
G:
W:
R:
3.50 mol NaOH
2 different substances
? g Na2SO4
You see g Na2SO4
2 mol NaOH = 1 mol Na2SO4
1 mol Na2SO4 = 142.042g Na2SO4
3.50 mol NaOH
1 mol Na2SO4
142.042g Na2SO4
x
x
2 mol NaOH
1 mol Na2SO4
1
Given over 1
Mole ratio
1 mol = 142.042g
= g Na2SO4
249
Actual Yield
Theoretical Yield
X 100
= % Yield
Same as…
% Yield
100
Actual Yield
Theoretical Yield
=
From the experiment
221 g NaSO4
249 g NaSO4
X 100
= 88.8%
From the calculation
87.3%
97.7%
3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas.
Zn + 2HCl H2 +ZnCl2
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Actual yield: 93.7 g
Actual from the experiment
2.75 g Cu
= 87.3 %
X 100
3.15 g Cu
calculated
A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)
Actual yield
actual
= % yield
X 100
Theo.
2Mg + O2 2 MgO
G:
W:
R:
5.00 g Mg
2 different substances
g MgO
You see g Mg and g MgO
2 mol Mg = 2 mol MgO
1 mol Mg = 24.305 g Mg
1 mol MgO = 40.304 g MgO
5.00 g Mg
1 mol Mg
40.304 g MgO
2 mol MgO
x
x
x
24.305 g Mg
2 mol Mg
1 mol MgO
1
8.29
Theoretical yield= g MgO
A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)
Actual yield
8.10 g MgO
X 100
= 97.7%
8.29 g MgO
3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas.
Zn + 2HCl H2 +ZnCl2
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Actual yield: 93.7 g
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.
Zn + 2HCl H2 +ZnCl2
G:
W:
R:
58.0 g HCl
2 different substances
g ZnCl2
You see g HCl & g ZnCl2
2 mol HCl = 1 mol ZnCl2
1 mol HCl = 36.461 g HCl
1 mol ZnCl2 = 136.286 g ZnCl2
58.0 g HCl
1 mol HCl
1 mol ZnCl2
136.286 g ZnCl2
x
x
x
36.461 g HCl
2 mol HCl
1 mol ZnCl2
1
108
Theoretical yield= g ZnCl2
Actual from the experiment
93.7 g
actual
= 87 %
X 100
108 g
calculated