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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 9: CHEMICAL COMPOSITION (PART 3)
CLASS NOTES
MOLE TO MOLE CONVERSIONS
Chemical equations are quantitative because they tell us how many reactants and products interact in a
given reaction. In particular, chemical reactions are written in mole to mole ratios. For example, 3
H2(g) + N2(g)  2 NH3(g) means that 3 moles of hydrogen gas react with 1 mole of nitrogen gas to
produce 2 moles of ammonia gas.
MOLE TO MOLE APPLICATIONS
Example 1. How many moles of water form from 24.6 moles oxygen in the following reaction:
O2(g) + 2 H2(g)  2 H2O(g)
1A.
According to the equation, 1 mole of oxygen reacts with excess hydrogen to produce 2 moles of water
vapor:
24.6 moles O2 (2 moles H2O)
-----------------1 mole O2
= 49.2 mol H2O
Example 2. How many moles of oxygen are needed to produce 425 moles of octane in the following
reaction?
__ C8H18(g) + __ O2(g)  __ H2O(l) + __ CO2(g)
2A.
(1) Balance the equation:
2 C8H18(g) + 25 O2(g)  18 H2O(l) + 16 CO2(g)
(2) 425 mol C8H18 (25 mol O2)
----------------2 mol C8H18
= 5.31 x 103 mol O2
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
MASS TO MASS CONVERSIONS
Mass to mass conversions are similar to mole to mole conversions. In general, they take the form:
Mass A  Moles A  Moles B  Mass B
Example 3. How many grams HCl (aq) are needed to consume 5.50 grams Mg(OH)2?
__ Mg(OH)2(aq) + __ HCl(aq)  __ H2O(l) + __ MgCl2(aq)
3A.
(1) Balance the equation:
(2)
(3)
(4)
(5)
(6)
(7)
Mg(OH)2(aq) + 2 HCl(aq) 
2 H2O(l) + MgCl2(aq)
Mass Mg(OH)2  Moles Mg(OH)2:
Mg = 24.31 amu
2 x O = 32.00 amu
2 x H = 2.02 amu
Mg(OH)2 = 58.33 g/mol
5.50 g Mg(OH)2 (1 mol)
---------58.33 g
= 9.43 x 10-2 mol Mg(OH)2
(8) Moles Mg(OH)2  Moles HCl:
(9) 9.43 x 10-2 mol Mg(OH)2
(2 mol HCl)
-------------------1 mol Mg(OH)2
= 1.89 x 10-1 mol HCl
(10)
(11)
(12)
(13)
H = 1.01 amu
Cl = 35.45 amu
HCl = 36.46 g/mol
Moles HCl  Mass HCl:
(14) 1.89 x 10-1 mol HCl
(36.46 g)
-------------1 mol HCl
= 6.88 g HCl
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 4: 2.6 x 103 kg SO2 reacts with excess oxygen and water. How many kilograms of H2SO4
does this produce?
__ SO2(g) + __ O2(g) + __ H2O(l)  __ H2SO4(aq)
4A.
(1) Balance the equation:
(2)
(3)
(4)
(5)
(6)
2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq)
Mass SO2  Moles SO2:
S = 32.07 amu
2 x O = 32.00 amu
SO2 = 64.07 g/mol
2.6 x 106 g SO2 (1 mol)
------------64.07 g SO2
= 4.05 x 104 mol SO2
(7) Moles SO2  Moles H2SO4:
(8) 4.05 x 104 mol SO2 (2 mol H2SO4)
-----------------1 mol SO2
= 8.12 x 104 mol H2SO4
(9) Moles H2SO4  Mass H2SO4:
(10) 2 x H = 2.02 amu
(11) S = 32.07 amu
(12) 4 x O = 64.00 amu
(13) H2SO4 = 98.09 g/mol
(14) 8.12 x 104 mol H2SO4
( 98.09 g)
---------------1 mol H2SO4
= 7.96 x 106 g  8.0 x 103 kg H2SO4
LIMITING REACTANT
Limiting reactants are compounds completely consumed in chemical reactions.
THEORETICAL YIELD
Theoretical yield is the amount of product that can be made in a chemical reaction. It is based on the
amount of limiting reagent.
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
ACTUAL YIELD
Actual yield is the amount of product actually produced by a chemical reaction.
PERCENTAGE YIELD
Percentage yield is a percentage based on the actual versus theoretical yields:
% Yield = (Actual Yield) / (Theoretical)
Example 5. 4.8 mol Na and 2.6 mol F2 are added to the following reaction. What is the limiting
reactant?
__ Na(s) + __ F2(g)  __ NaF(s)
5A.
(1) Balance the equation:
2 Na(s) + F2(g)  2 NaF(s)
Theoretical product yield from Na:
(2) 4.8 mol Na (2 mol NaF) = 4.8 mol NaF
---------------2 mol Na
Theoretical product yield from F2:
(3) 2.6 mol F2 (2 mol NaF) = 5.2 mol NaF
--------------1 mol F2
(4) Limiting reactant = Na (4.8 mol NaF)
Example 6. 185 grams Fe2O3 and 95.3 grams CO produce 87.4 grams Fe(s) in a laboratory.
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
6a. What is the limiting reactant?
A.
Mass Fe2O3  Moles Fe2O3:
(1) 2 x Fe = 111.7 amu
(2) 3 x O = 48.00 amu
(3) Fe2O3 = 159.7 g/mol
(4) 185 g Fe2O3 (1 mol) = 1.16 mol Fe2O3
---------159.7 g
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CHEMISTRY
MR. SURRETTE
VAN NUYS HIGH SCHOOL
6a. (continued…)
Moles Fe2O3  Moles Fe:
(5) 1.16 mol Fe2O3
(2 mol Fe)
---------------1 mol Fe2O3
= 2.32 mol Fe
From Fe2O3: 2.32 mol Fe
Mass CO  Moles CO:
(6) C = 12.01 amu
(7) O = 16.00 amu
(8) CO = 28.01 g/mol
(9) 95.3 g CO
(1 mol) = 3.40 mol CO
----------28.01 g
Moles CO  Moles Fe:
(10) 3.40 mol CO (2 mol Fe) = 2.27 mol Fe
------------3 mol CO
(11) From CO: 2.27 mol Fe
(12) 2.27 mol Fe < 2.32 mol Fe
(13) Limiting reactant = CO
6b. What is the theoretical yield of solid iron?
A.
(1) Theoretical yield = product from limiting reactant
(2) 2.27 mol Fe (55.75 g) = 126.4 g Fe
-----------1 mol Fe
(3) Theoretical yield = 126.4 g Fe
6c.
A.
(1)
(2)
(3)
(4)
(5)
What is the percent yield for this reaction?
% Yield = Actual Yield / Theoretical Yield
Actual yield = 87.4 g Fe
Theoretical yield = 126.4 g Fe
% Yield = 87.4 g Fe / 126.4 g Fe
Yield = 69.1%
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CHEMISTRY