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Free Group Study Tutoring
• Monday 5-6pm • Tuesday 5-6pm• Friday 11am-12pm• Tutoring will begin Friday Sept. 16th at
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562-985-5350
Stoichiometry
stoi·chi·om·e·try nounLiterally – measuring the componentsCalculations of the quantitative relationships between reactants and products in a chemical reaction.
The Mole and Chemical Reactions:The Macro-Nano Connection
2 H2 + O2 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules
2 moles H2 molecules 1 mole O2 molecules 2 moles H2O molecules
4.0 g H2 32.0 g O2 36.0 g H2O
EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?
Assume that the rail is 132 lb Fe/yard and that the weld is 1/10 inch wide.
weld
Photo by Mike Condren
The mass of iron in 1/10 inch of this rail is:
g166lb1
g454
yd1
lb132
in36
yd1in10
1
Fe2O3 + 2 Al Al2O3 +2 Fe
3232
3232 OFeg237OFemol1
OFeg68.159
Femol2
OFemol1
Feg84.55
Femol1Feg166
Alg.80Almol1
Alg98.26
Femol2
Almol2
Feg84.55
Femol1Feg166
EXAMPLE What is the number of moles of CaSO4 (S) that can be produced by allowing 1.0 mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react?
2SO2(g) + 2CaCO3(s) + O2(g) 2CaSO4(S) + 2CO2(g)
balanced equation relates:2SO2(g) 2CaCO3(s) O2(g)
have only:1SO2(g) 2CaCO3(s) 3O2(g)
not enough SO2 to use all of the CaCO3 or the O2
not enough CaCO3 to use all of the O2
SO2 is the limiting reactant
Compare Amounts of Product
42
42 CaSOmol1
SOmol2
CaSOmol2SOmol1
43
43 CaSOmol2
CaCOmol2
CaSOmol2CaCOmol2
42
42 CaSOmol6
Omol1
CaSOmol2Omol3
SO2 limits the amount of product formed
Mass of Product
• 95.0 g of chlorine and 27.0 g of phosphorus react to form PCl3. What mass of PCl3 is formed?
P4 (s) + 6 Cl2 (g) 4 PCl3 (l)
33
3
2
3
2
22 PClg7.122
PClmol1
PClg32.137
Clmol6
PClmol4
Clg90.70
Clmol1Clg0.95
33
3
4
3
4
44 PClg7.119
PClmol1
PClg32.137
Pmol1
PClmol4
Pg88.123
Pmol1Pg0.27
27.0g phosphorus gives the smaller yield. Phosphorus is the limiting reactant it limits the amount of product formed
Excess Reactant
• We can also calculate the amount of the non-limiting reactant that is used up, and thus how much is left over
23
2
4
2
4
44 Clg72.92
PClmol1
Clg90.70
Pmol1
Clmol6
Pg88.123
Pmol1Pg0.27
92.72 g Cl2 are required to react with all the P4
Thus 95.0-92.72=2.3 g Cl2 are left over as excess
Theoretical Yield
The amount of product produced by a reaction based on the amount of the limiting reactant
Percent Yield
actual yield% yield = 100% theoretical yield
For example, if only 103.5 g of PCl3
were actually produced
yield%5.86%100PClg7.119
PClg5.103yield%
3
3
EXAMPLE A rocket fuel, hydrazine, is produced by a reaction of Cl2 with excess NaOH and NH3. (a) What theoretical yield can be produced from 1.00 kg of Cl2?
2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for the overall process
(1.00 kg Cl2) (1 kmol Cl2)
(70.9 kg Cl2) molar mass
(1 kmol N2H4)
(1 kmol Cl2)balancedequation
(32.0 kg N2H4)
(1 kmol N2H4)molar mass
= 0.451 kg N2H4
EXAMPLE (b) What is the actual yield if 0.299 kg of 98.0% N2H4 is produced for every 1.00 kg of Cl2?
(98.0 kg N2H4)
(100 kg product)purity factor
(0.299 kg product) = 0.293 kg N2H4
2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4
(b) actual yield
EXAMPLE (c) What is the percent yield of pure N2H4?
2NaOH + Cl2 + 2NH3 N2H4 + 2NaCl + 2H2O(a) theoretical yield = 0.451 kg N2H4
(b) actual yield = 0.293 kg N2H4
(c) percent yield 0.293 kg
% yield = 100% = 65.0 % yield 0.451kg
Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO2 and 2.85-mg of H2O formed. What is the empirical formula of benzoic acid?
(16.4-mg of CO2 )(12.01-mg C) # mg C = = 4.48-mg C (44.01-mg CO2)
(2.85-mg of H2O )(2.01-mg H) # mg H = = 0.318-mg H (18.01-mg H2O)