Download Ch. 9 Review

CH. 9
review
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Ch. 9 Review
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Terms/definitions
Composition
stoichiometry
Reaction
stoichiometry
Mole ratio
Molar mass
Theoretical yield
Actual yield
Percent yield
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Problems
Moles to moles
Mass to moles
Moles to mass
Mass to mass
Percent yield
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Matching
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DEALS WITH THE MASS
RELATIONSHIPS OF ELEMENTS IN
COMPOUNDS
INVOLVES THE MASS RELATIONSHIPS
BETWEEN REACTANTS AND PRODUCTS
IN A CHEMICAL REACTION
A CONVERSION FACTOR THAT RELATES
THE AMOUNTS IN MOLES OF ANY TWO
SUBSTANCES INVOLVED IN A CHEMICAL
REACTION
MASS, IN GRAMS, OF ONE MOLE OF A
SUBSTANCE
THE MAXIMUM AMOUNT OF PRODUCT
THAT CAN BE PRODUCED FROM A GIVEN
AMOUNT OF REACTANT (RESULT OF
MASS-MASS PROBLEM)
THE MEASURED AMOUNT OF A PRODUCT
OBTAINED FROM A REACTION
RATIO OF ACTUAL YIELD TO THE
THEORETICAL YIELD, MULTIPLIED BY
100
mole ratio
actual yield
molar mass
composition
stoichiometry
percent
yield
reaction
stoichiometry
theoretical
yield
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Ch. 9 Test: 45 points
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7 matching: terms/def.
2 multiple choice: coefficients & mole ratio
8 problems
moles to moles (2)
 moles to grams
 grams to moles
 grams to grams (2)
 percent yield with grams to grams
 percent yield, with grams to grams, find
actual yield given percent yield
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Practice all 4 types
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Do p. 287 Section Review # 2 – 5
2a) 182 g HCl b) 238 g MgCl2; 5.05 g H2
3a) 0.998 mol H2O
3b) 0.499 mol C2H2; 0.499 mol Ca(OH)2
4) NaCl + AgNO3  AgCl + NaNO3
4) 63.3 g AgCl
5) 8.45 x 104 g CO2; 1.73 x 104 g H2O
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Review Problems pp. 930 - 935
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Mixed Stoichiometry problems # 176 – 199
176) 15 mol (NH4)2SO4
177a) 51 g Al
b) 101 g Fe
177c) 1.83 mol Fe2O3
178) 0.303 g H2
179) H2SO4 + 2 KOH  K2SO4 + 2 H2O
1.11 g H2SO4
180a) H3PO4 + 2 NH3  (NH4)2HPO4
b) 0.293 mol (NH4)2HPO4
c) 970 kg NH3 = 970 000 g NH3
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Review Problems pp. 930 - 935
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Mixed Stoichiometry problems # 176 – 199
181a) 90.0 mol ZnCO3; 60.0 mol C6H8O7
b) 13.5 kg (13500 g) H2O;
33.0 kg (33 000 g) CO2
182a) 60.9 g methyl butanoate
b) 3261 g H2O
183a) 0.450 mol N2 b) 294 g NH4NO3
184) Pb(NO3)2 + 2 KI  PbI2 + 2 KNO3
0.751 mg KNO3 (0.000751 g)
185) 3.3 mol PbSO4
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Review Problems pp. 930 - 935
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Mixed Stoichiometry problems # 176 – 199
186) 2 LiOH + CO2  H2O + Li2CO3
360 g H2O
187a) 38.1 g H2O b) 40.1 g H3PO4
c) 0.392 mol H2O
188) C2H5OH + 3 O2  2 CO2 + 3 H2O
81.0 g C2H5OH
189) 76.5 g H2SO4; 12.5 g O2
190) 2 NaHCO3  Na2CO3 + H2O + CO2
1.31 g CO2
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Review Problems pp. 930 - 935
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Mixed Stoichiometry problems # 176 – 199
191a) 2 N2H4 + N2O4  3 N2 + 4 H2O
b) 1 mol N2O4 to 3 mol N2
c) 30 000 mol N2 d) 3.52 x 105 g H2O
192) 2 HgO  2 Hg + O2
1.1954 mol O2
193) 2 Fe + 3 Cl2  2 FeCl3
30.0 g Fe
194) 9.26 mg CdS (0.00926 g)
195a) 1.59 mol CO2 b) 0.0723 mol C3H5(OH)3
c) 535 g Mn2O3 d) 833 g C3H5(OH)3
4.97 g CO2
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Review Problems pp. 930 - 935
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Mixed Stoichiometry problems # 176 – 199
196a) 3.29 x 103 kg of HCl (329 000 000 g)
b) 330 g CO2
197a) 6.53 x 105 g NH4ClO4
b) 160 kg NO (160 000 g)
198a) 1.70 x 106 mol H3PO4
b) 666 kg CaSO4 . 2 H2O (666 000 g)
c) 34 metric tons of H3PO4
199) 1670 kg (1 670 000 g)
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Review Problems pp. 930 - 935
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Percent yield # 218 – 236
218a) 83.8 % b) 93 % c) 69.1 %
d) 46 %
219a) 79.3 % b) 76 %
220a) 64.3 % b) 58.0 % c) 69.l5 %
skip d
221a) 69.5 % b) 79.0 % c) 48 %
d) 85 %
222a) 59 % b) 81.0 % c) 2.3 x 105 mol P
223a) 91.8 % b) 0.0148 mol W
c) 16.1 g WO3
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Review Problems pp. 930 - 935
Percent yield # 218 – 236
 224a) 86.8 % b) 92.2 % skip c
 225a) 81 % b) 2.0 x 102 g N2O5
 226) 80.1 %
 227a) 95 %
b) 910 g Au skip c
 228a) 87.5 % b)0.25 g CO
 229a) 71 % b) 26 metric tons
c) 47.8 g NaCl skip d
230a) 2 Mg + O2  2 MgO
b) 87.7 %
c) 3 Mg + N2  Mg3N2
d) 84 %
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Review Problems pp. 930 - 935
Percent yield # 218 – 236
 231a) 80 % b) 66.2 % c) 57.1 %
 232) 2 C3H6 + 2 NH3 + 3 O2 
2 C3H3N + 6 H2O
91 %
 233a) CO + 2 H2  CH3OH
3410 kg or 3 410 000 g
b) 91.5 %
234) 96.9 %
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Review Problems pp. 930 - 935
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Percent yield # 218 – 236
235) 6 CO2 + 6 H2O  C6H12O6 + 6 O2
6320 g O2
236) 27.6 kg O2 = 27 600 g O2
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