stoker exercises
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How is stoichiometry like baking cookies?. Stoker Exercises. A recipe indicates the amount of each ingredient and the procedure used to produce a certain number of cookies. By looking at the list of ingredients, you can predict what the finished cookie will be like. ?. Stoichiometry. - PowerPoint PPT PresentationTRANSCRIPT
Stoker Exercises
How is stoichiometry like baking cookies?
A recipe indicates the amount of each ingredient and the procedure used to produce a certain number of cookies. By looking at the list of ingredients, you can predict what the finished cookie will be like.
?
10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant?
a) SiO2 + 3C 2CO + SiC
X ------------------ X ------------------ X ------------------
60.09 g SiO2
1 mol SiO2
1 mol SiO2
3 mol C
1 mol C
12.01 g C
g C = ? 1.772 g SiO2
= 1.062 g C
Stoichiometry
10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant?
b) 5 O2 + C3H8 3CO2 + 4H2O
X ------------------ X ------------------ X ------------------
32.00 g O2
1 mol O2
5 mol O2
1 mol C3H8
1 mol C3H8
44.09 g C3H8
g C3H8= ? 1.772 g O2
= 0.4883 g C3H8
Stoichiometry
10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant?
c) CH4 + 4Cl2 4HCl + CCl4
X ------------------ X ------------------ X ------------------
16.04 g CH4
1 mol CH4
1 mol CH4
4 mol Cl2
1 mol Cl2
70.90 g Cl2
g Cl2= ? 1.772 g CH4
= 31.33 g Cl2
Stoichiometry
10.45 How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 1.772 g of the first-listed reactant?
d) 3NO2 + H2O 2HNO3 + NO
X ------------------ X ------------------ X ------------------
46.01 g NO2
1 mol NO2
3 mol NO2
1 mol H2O
1 mol H2O
18.02 g H2O
g H2O= ? 1.772 g NO2
= 0.2313 g H2O
Stoichiometry
10.46 a How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant?
a) H2O2 + H2S 2H2O + S
X ------------------ X ------------------ X ------------------
34.02 g H2O2
1 mol H2O2
1 mol H2O2
1 mol H2S
1 mol H2S
34.09 g H2S
g H2S = ? 12.56 g H2O2
= 12.59 g H2S
Stoichiometry
10.46 b How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant?
b) 4NH3 + 3O2 2N2 + 6H2O
X ------------------ X ------------------ X ------------------
17.04 g NH3
1 mol NH3
4 mol NH3
3 mol O2
1 mol O2
32.00 g O2
g O2 = ? 12.56 g NH3
= 17.69 g O2
Stoichiometry
10.46 c How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant?
c) Mg + 2HCl MgCl2 + H2
X ------------------ X ------------------ X ------------------
24.30 g Mg
1 mol Mg
1 mol Mg
2 mol HCl
1 mol HCl
36.46 g HCl
g HCl = ? 12.56 g Mg
= 37.69 g HCl
Stoichiometry
10.46 d How many grams of the second-listed reactant in each of the following reactions is needed to react completely with 12.56 g of the first-listed reactant?
d) 6HCl + 2Al 3H2 + 2AlCl3
X ------------------ X ------------------ X ------------------
36.46 g HCl
1 mol HCl
6 mol HCl
2 mol Al
1 mol Al
26.98 g Al
g Al = ? 12.56 g HCl
= 3.098 g Al
Stoichiometry
10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s) SiC(s) + 2CO(g)
a) How many grams of SiO2 are needed to react with 1.50 moles of C?
a) SiO2 + 3C SiC + 2CO
X ------------------ X ------------------
3 mol C
1 mol SiO2
1 mol SiO2
60.1 g SiO2
g SiO2 = ? 1.50 mol C
= 30.1 g SiO2
Stoichiometry
10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s) SiC(s) + 2CO(g)
b) How many grams of CO are produced when 1.37 moles of SiO2 react?
b) SiO2 + 3C SiC + 2CO
X ------------------ X ------------------
1 mol SiO2
2 mol CO
1 mol CO
28.0 g CO
g CO = ? 1.37 mol SiO2
= 76.7 g CO
Stoichiometry
10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s) SiC(s) + 2CO(g)
c) How many grams of SiC are produced at the same time that 3.33 moles of CO are produced?
c) SiO2 + 3C SiC + 2CO
X ------------------ X ------------------
2 mol CO
1 mol SiC
1 mol SiC
40.1 SiC
g SiC = ? 3.33 mol CO
= 66.8 g SiC
Stoichiometry
10.47 Silicon carbide, SiC, used as an abrasive on sandpaper, is prepared using the following chemical reaction: SiO2(s) + 3C(s) SiC(s) + 2CO(g)
d) How many grams of C must react in order to produce 0.575 mole of SiC?
d) SiO2 + 3C SiC + 2CO
X ------------------ X ------------------
1 mol SiC
3 mol C
1 mol C
12.0 g C
g C = ? 0.575 mol SiC
= 20.7 g C
Stoichiometry
10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s) 2 Na(s) + 3N2(g) a) How many grams of NaN3 must decompose to produce 3.57 moles of N2?
a) 2NaN3 2 Na + 3N2
X ------------------ X ------------------
3 mol N2
2 mol NaN3
1 mol NaN3
65.0 NaN3
g NaN3 = ? 3.57 mol N2
= 155 g NaN3
Stoichiometry
10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s) 2 Na(s) + 3N2(g) b) How many grams of NaN3 must decompose to produce 3.57 moles of Na?
b) 2NaN3 2 Na + 3N2
X ------------------ X ------------------
2 mol Na
2 mol NaN3
1 mol NaN3
65.0 NaN3
g NaN3 = ? 3.57 mol Na
= 232 g NaN3
Stoichiometry
10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s) 2 Na(s) + 3N2(g) c) How many grams of Na are produced at the same time that 5.40 moles of N2 are produced?
c) 2NaN3 2 Na 3N2
X ------------------ X ------------------
3 mol N2
2 mol Na
1 mol Na
23.0 Na
g Na = ? 5.40 mol N2
= 82.8 g Na
Stoichiometry
10.48 The inflating gas for automobile air bags is nitrogen (N2), generated from the decomposition of sodium azide (NaN3). The equation for the decomposition reaction is: 2NaN3(s) 2 Na(s) + 3N2(g) d) How many moles of NaN3 must decompose in order to produce 10.00 g of N2?
d) 2NaN3 2 Na 3N2
X ------------------ X ------------------
28.02 g N2
1 mol N2
3 mol N2
2 mol NaN3
mol NaN3 = ? 10.00 g N2
= 0.2379 mol NaN3
Stoichiometry
2LiOH + CO2 Li2CO3 + H2O
X ------------------ X ------------------
1 mol CO2
2 mol LiOH
1 mol LiOH
23.9 g LiOH
g LiOH = ? 4.50 mol CO2
215 g LiOH
10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: a) absorb 4.50 moles of CO2
Stoichiometry
2LiOH + CO2 Li2CO3 + H2O
X ------------------ X ------------------
6.02 X 1023 molec. CO2
1 mol CO2
1 mol CO2
2 mole LiOH
g LiOH = ? 3.00 X 1024 molc. CO2
238 g LiOH
10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: b) absorb 3.00 X 1024 molec. CO2
1 mol LiOH
23.9 g LiOH
X ------------------
Stoichiometry
2LiOH + CO2 Li2CO3 + H2O
X ------------------ X ------------------
18.0 g H2O
1 mol H2O
1 mol H2O
2 mole LiOH
g LiOH = ? 10.0 g H2O
26.6 g LiOH
10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: c) produce 10.0 g H2O
X ------------------
1 mol LiOH
23.9 g LiOH
Stoichiometry
2LiOH + CO2 Li2CO3 + H2O
X ------------------ X ------------------
73.9 g Li2CO3
1 mol Li2CO3
1 mol Li2CO3
2 mole LiOH
g LiOH = ? 10.0 g Li2CO3
6.47 g LiOH
10.49 One way to remove gaseous carbon dioxide (CO2) from the air in a spacecraft is to let canisters of solid lithium hydroxide (LiOH) absorb it according to the reaction 2LiOH + CO2 Li2CO3 + H2O. Based on this equation, calculate how many grams of LiOH must be used to achieve the following: d) produce 10.0 g of Li2CO3
X ------------------
1 mol LiOH
23.9 g LiOH
Stoichiometry
10.50 a) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g) W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O
b) 2.53 moles of H2O d) 250 000 atoms of W
X ------------------ X ------------------
183.8 g W
1 mol W
1 mol W
1 mol WO3
g WO3 = ? 10.00 g W
X ------------------
1 mol WO3
231.8 g WO3
WO3(s) + 3H2(g) W(s) + 3H2O(l)
12.61 g WO3
Stoichiometry
10.50 b) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g) W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O
c) 2.53 moles of H2O d) 250 000 atoms of W
X ------------------ X ------------------
6.02 X 1023 molec. H2O
1 mol H2O
3 mol H2O
1 mol WO3
g WO3 = ? 1.00 X 109 molec. H2O
X ------------------
1 mol WO3
231.8 g WO3
WO3(s) + 3H2(g) W(s) + 3H2O(l)
1.28 X 10-13 g WO3
Stoichiometry
10.50 c) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g) W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O
c) 2.53 moles of H2O d) 250 000 atoms of W
X ------------------ X ------------------
3 mol H2O
1 mol WO3
1 mol WO3
231.8 g WO3
g WO3 = ? 2.53 mol H2O
WO3(s) + 3H2(g) W(s) + 3H2O(l)
= 195 g WO3
Stoichiometry
10.50 d) Tungsten (W) metal, used to make incandescent light bulb filaments, is produced by the reaction WO3(s) + 3H2(g) W(s) + 3H2O(l). Based on this equation, calculate how many grams of WO3 are needed to produce each of the following: a) 10.00 g of W b) 1.00 X 109 molec. H2O
c) 2.53 moles of H2O d) 250 000 atoms of W
X ------------------ X ------------------
6.0 X 1023 atoms W
1 mol W
1 mol W
1 mol WO3
g WO3 = ? 2.5 X 105
atoms W
WO3(s) + 3H2(g) W(s) + 3H2O(l)
= 9.7 X 10 -17 g WO3
X ------------------
1 mol WO3
231.8 g WO3
Stoichiometry
Stoichiometry
Stoichiometry
X ------------------ X ------------------
36.46 g HCl
1 mol HCl
8 mol HCl
1 mol K2S2O3
mol K2S2O3 = ? 2.500 g HCl
= 0.008571 mol K2S2O
K2S2O3 + 4Cl2 + 5H2O 2 KHSO4 + 8HCl
Stoichiometry
Stoichiometry
Stoichiometry
Stoichiometry
10.53
g Na = ? 16.5 g S
How many grams of sodium are needed to react completely with 16.5 g of sulfur in the synthesis of Na2S?
2Na + S Na2S
16.5 g S X -------------
32.1 g S
1 mol SX -------------
1 mol S
2 mol NaX -------------
1 mol Na
23.0 g Na
= 23.6 g Na
Stoichiometry
10.54
g Be = ? 45.0 g N2
How many grams of beryllium are needed to react completely with 45.0 g of nitrogen in the synthesis of Be3N2?
3Be + N2 Be3N2
45.0 g N2X -------------
28.0 g N2
1 mol N2X -------------
1 mol N2
3 mol BeX -------------
1 mol Be
9.01 g Be
= 43.4 g Be
Stoichiometry
10.55
g Cr = ? 200.0 g CrCl3
When chromium metal reacts with chlorine gas, a violet solid with the formula CrCl3 is formed: 2Cr + 3Cl2 2CrCl3. How many grams of Cr and how many grams of Cl2 are needed to produce 200.0 g of CrCl3?
Mass of Chromium:
200.0 g CrCl3 X ------------- 158.4 g CrCl3
1 mol CrCl3X -------------2 mol CrCl3
2 mol CrX -------------1 mol Cr
52.00 g Cr
= 65.66 g Cr
Stoichiometry
10.55
g Cl2 = ? 200.0 g CrCl3
When chromium metal reacts with chlorine gas, a violet solid with the formula CrCl3 is formed: 2Cr + 3Cl2 2CrCl3. How many grams of Cr and how many grams of Cl2 are needed to produce 200.0 g of CrCl3?
Mass of Chlorine:
200.0 g CrCl3 X ------------- 158.4 g CrCl3
1 mol CrCl3X -------------2 mol CrCl3
3 mol Cl2X -------------1 mol Cl2
70.90 g Cl2
= 134.3 g Cl2
Stoichiometry
g Ag = ? 150.0 g Ag2S
150.0 g Ag2S X ------------- 247.9 g Ag2S
1 mol Ag2SX -------------1 mol Ag2S
2 mol AgX -------------1 mol Ag
107.9 g Ag
=130.6 g Ag
10.56 Black silver sulfide can be produced from the reaction of silver metal with sulfur: 2Ag + S Ag2S. How many grams of Ag and how many grams of S are needed to produce 150.0 g of Ag2S?
Part 1
Stoichiometry
g S = ? 150.0 g Ag2S
10.56 Black silver sulfide can be produced from the reaction of silver metal with sulfur: 2Ag + S Ag2S. How many grams of Ag and how many grams of S are needed to produce 150.0 g of Ag2S?
150.0 g Ag2S X ------------- 247.9 g Ag2S
1 mol Ag2SX -------------1 mol Ag2S
1 mol SX -------------1 mol S
32.07 g S
=19.41 g S
Part 2
Stoichiometry
10.57
Combinations = ? 216 nuts 284 bolts
What will be the limiting reactant in the production of “three nut-four bolt” combinations from a collection of 216 nuts and 284 bolts?
216 nuts X -------------
3 nuts
1 combination= 72 combinations
284 bolts X -------------
4 bolts
1 combination= 71 combinations
The limiting reactant is 284 bolts
Limiting Reactant
Nuts is the limiting reactant
10.58 What will be the limiting reactant in the production of “five nut-four bolt” combinations from a collection of 785 nuts and 660 bolts?
Combinations = ? 785 nuts 660 bolts
785 nuts X -----------------
1 combination
5 nuts
= 157 combinations
660 bolts X -----------------
1 combination
4 bolts
= 165 combinations
Limiting Reactant
10.59 A model airplane kit is designed to contain two wings, one fuselage, four engines, and six wheels. How many model airplane kits can a manufacturer produce from a parts inventory of 426 wings, 224 fuselages, 860 engines, and 1578 wheels?
kits = ? 426 wings, 224 fuselages, 860 engines, and 1578 wheels
426 wings X -----------------1 kit
2 wings= 213 kits
224 fuselages X -----------------1 kit
1 fuselage
= 224 kits
860 engines X -----------------1 kit
4 engines
=215 kits
1578 wheels X -----------------1 kit
6 wheels
= 263 kits
Limiting Reactant
10.60 A model car kit is designed to contain one body, four wheels, two bumpers, and one steering wheel. How may model car kits can a manufacturer produce from a parts inventory of 137 bodies, 532 wheels, 246 bumpers, and 139 steering wheels?
kits = ? 137 bodies, 532 wheels, 246 bumpers, and 139 steering wheels
137 bodies X -----------------1 kit
1 body= 137 kits
532 wheels X -----------------1 kit
4 wheels
= 133 kits
246 bumpers X -----------------1 kit
2 bumpers
=123 kits
139 steering wheels X -----------------1 kit
1 steering wheel
= 139 kits
Limiting Reactant
10.61 At high temperatures and pressures nitrogen will react with hydrogen to produce ammonia as shown by the equation N2 + 3H2 2NH3. For each of the following combinations of reactants, decide which is the limiting reactant.
a) 1.25 mol N2 and 3.65 mol H2 b) 2.60 mol N2 and 8.00 mol H2
c) 44.0 N2 and 3.00 mol H2 d) 55.0 g N2 and 15.0 g H2
Limiting reactant = ? 1.25 mol N2 3.65 mol H2
1.25 mole N2 X -----------------2 mol NH3
1 mol N2
= 2.50 mol NH3
3.65 mol H2 X -----------------2 mol NH3
3 mol H2
=2.43 mol NH3
3.65 mol H2 is the limiting reactant.
10.61 a) 1.25 mol of N2 and 3.65 mol H2
N2 + 3H2 2NH3Limiting Reactant
Limiting reactant = ? 2.60 mol of N2 and 8.00 mol H2
2.60 mole N2 X -----------------2 mol NH3
1 mol N2
= 5.20 mol NH3
8.00 mol H2 X -----------------2 mol NH3
3 mol H2
=5.33 mol NH3
2.60 mol N2 is the limiting reactant.
10.61 b) 2.60 mol of N2 and 8.00 mol H2
N2 + 3H2 2NH3Limiting Reactant
Limiting reactant = ? 44.0 g of N2 and 3.00 mol H2
44.0 g N2 X -----------------1 mol N2
28.0 g N2
= 3.14 mol NH3
3.00 mol H2 X -----------------2 mol NH3
3 mol H2
= 2.00 mol NH3
3.00 mol H2 is the limiting reactant.
10.61 c) 44.0 g of N2 and 3.00 mol H2
X -----------------
1 mol N2
2 mol NH3
N2 + 3H2 2NH3
Limiting reactant = ? 44.0 g of N2 and 3.00 mol H2
55.0 g N2 X -----------------1 mol NH3
28.0 g N2
= 3.93 mol NH3
3.00 mol H2 X -----------------1 mol NH3
2.02 g H2
= 4.95 mol NH3
55.0 g N2 is the limiting reactant.
10.61 d) 55.0 g of N2 and 15.0 g H2
X -----------------
1 mol N2
2 mol NH3
N2 + 3H2 2NH3
X -----------------
3 mol H2
2 mol NH3
4Al + 3O2 2Al2O3
Limiting Reactant
Limiting reactant = ? 3.00 mol Al and 4 mol O2
3.00 mol Al X -----------------2 mol Al2O3
4 mol Al
=1.50 mol Al2O3
4.00 mol O2 X -----------------2 mol Al2O3
3 mol O2
= 2.67 mol Al2O3
3.00 mol Al is the limiting reactant.
10.6 a) 3.00 mol Al and 4.00 mol O2
4Al + 3O2 2Al2O3Limiting Reactant
Limiting reactant = ? 7.00 mol Al and 7.40 mol O2
7.00 mol Al X -----------------2 mol Al2O3
4 mol Al
=3.50 mol Al2O3
5.40 mol O2 X -----------------2 mol Al2O3
3 mol O2
= 3.60 mol Al2O3
7.00 mol Al is the limiting reactant.
10.6 b) 7.00 mol Al and 5.40 mol O2
4Al + 3O2 2Al2O3
Limiting reactant = ? 16.2 g Al and 0.40 mol O2
16.2 AlX -----------------
2 mol Al2O3
4 mol Al= 0.300 mol Al2O3
0.40 mol O2 X -----------------2 mol Al2O3
3 mol O2
= 0.27 mol Al2O3
0.40 mol O2 is the limiting reactant.
10.6 c) 16.2 g Al and 0.40 mol O2
4Al + 3O2 2Al2O3
27.0 g Al
1 mol AlX -----------------
Limiting reactant = ? 100.0 g Al and 100.0 g O2
100.0 g AlX -----------------
2 mol Al2O3
4 mol Al= 1.852 mol Al2O3
100.0 g O2 X -----------------1 mol Al2O3
32.00 g O2
= 2.083 mol Al2O3
100.0 g Al is the limiting reactant.
10.6 d) 100.0 g Al and 100.0 g O2
4Al + 3O2 2Al2O3
27.00 g Al
1 mol AlX -----------------
X -----------------
3 mol O2
2 mol Al2O3
How many grams of magnesium nitride can be produced from the following amounts of reactants?
a) 10.0 g of Mg and 10.0 g of N2
Limiting Reactant
10.63 a)
Determine the limiting reactant in each part, then convert moles of product to mass of product.
How many grams of magnesium nitride can be produced from the following amounts of reactants?
a) 10.0 g of Mg and 10.0 g of N2
g of Mg3N2 = ? 10.0 g of Mg and 10.0 g of N2
10.0g Mg X ----------------24.3 g Mg
1 mol Mg
X ----------------
3 mol Mg
1 mol Mg3N2
= 0.137 ml Mg3N2
28.0 g N2
1 mol N2
X ----------------
1 mol N2
1 mol Mg3N210.0 g N2 X --------------
= 0.357 mol Mg3N2
The Mg is the limiting reactant.
0.137 mol Mg3N2 X -------------------
1 mol Mg3N2
100.9 g Mg3N2 = 13.8 g Mg3N2
Limiting Reactant
10.63 b)
20.0 g Mg X ---------------------------
1 mol Mg
24.3 g Mg
X ------------------------
3 mol Mg
1 mol Mg3N2
= 0.274 mol Mg3N2
28.0 g N2
1 mol N2
X ----------------
1 mol N2
1 mol Mg3N210.0 g N2 X --------------
= 0.357 mol Mg3N2
0.274 mol Mg3N2 X -----------------------100.9 g Mg3N2
1 mol Mg3N2
= 27.6 g Mg3N2
Mg is the limiting reactant.
Limiting Reactant
10.63 c)
30.0 g Mg X ---------------------------
1 mol Mg
24.3 g Mg
X ------------------------
3 mol Mg
1 mol Mg3N2
= 0.412 mol Mg3N2
28.0 g N2
1 mol N2
X ----------------
1 mol N2
1 mol Mg3N210.0 g N2 X --------------
= 0.357 mol Mg3N2
0.357 mol Mg3N2 X -----------------------100.9 g Mg3N2
1 mol Mg3N2
= 36.0 g Mg3N2
N2 is the limiting reactant.
Limiting Reactant
10.63 d)
40.0 g Mg X ---------------------------
1 mol Mg
24.3 g Mg
X ------------------------
3 mol Mg
1 mol Mg3N2
= 0.549 mol Mg3N2
28.0 g N2
1 mol N2
X ----------------
1 mol N2
1 mol Mg3N210.0 g N2 X --------------
= 0.357 mol Mg3N2
0.357 mol Mg3N2 X -----------------------100.9 g Mg3N2
1 mol Mg3N2
= 36.0 g Mg3N2
N2 is the limiting reactant.
Limiting Reactant
70.0 g Fe3O4
12.0 g of oxygen gas sealed in a glass vessel.
Unreacted reactant = ? 70.0 g Fe3O4 12 g O2
The limiting reactant will all react. Use the limiting reactant to determine the mass of the other reactant that reacts, and then find the unreacted by difference.
Limiting Reactant
0.455 mol Fe2O3 2.25 mol Fe2O3
Iron(III) oxide
70.0 g Fe3O4X ---------------------------
1 mol Fe3O4
231. g Fe3O4
X ------------------------
4 mol Fe3O4
6 mole Fe2O3
= 0.455 mole Fe2O3
32.0 g O2
1 mol O2
X ----------------
1 mol O2
6 mol Fe2O312.0 g O2 X --------------
= 2.25 mol Fe2O3
The Fe3O4 is the limiting reactant. There will be none left upon completion. Calculate the mass of O2 reacted:
Limiting Reactant
0.455 mol Fe2O3 X ---------------------------
1 mol O2
6 mol Fe2O3
X ------------------------
1 mol O2
32.0 g O2
= 2.43 g O2
Limiting Reactant
Limiting Reactant
16.0 10.1
Limiting Reactant
g of products = ? 8.00 g SCl2 and 4.00 g NaF
g of Products = ? 100.00 g Fe3Br8
300.0 g Fe3Br8
Na2CO3 is the limiting reactant.
Limiting Reactant10.70 Determine the number of grams of each of the products that can be made from 100.00 g Na2CO3 and 300.0 g Fe3Br8by the following reaction. 4Na2CO3+ Fe3Br8 8NaBr + 4CO2 + Fe3O4
Calculation of Theoretical YieldThe theoretical yield of a reaction is the amount of product that would be formed if the reaction went to completion. It is based on the stoichiometry of the reaction and ideal conditions in which starting material is completely consumed, undesired side reactions do not occur, the reverse reaction does not occur, and there no losses in the work-up procedure.
Theoretical YieldActual Yield and Percent Yield
After your laboratory reaction is complete, you will isolate and measure the amount of product, then compare the actual yield to the theoretical yield to determine the percent yield:
Actual yield (in grams)-------------------------------- = X 100% = percent yield Theoretical yield (in grams)
Theoretical YieldActual Yield and Percent Yield
(a) What is the theoretical yield of NaClO that can be obtained from a reaction mixture containing 75.0 g of NaOH and 50.0g of Cl2?
(b) If the actual yield of NaClO for the reaction mixture in part (a) is 43.2 g, what is the percent yield of NaClO for the reaction?
Example 10.13The active ingredient in household laundry bleaches is sodium hypochlorite (NaClO).2NaOH + Cl2 NaCl + NaClO + H2O
Theoretical YieldActual Yield and Percent Yield
Answer for Example 10.13(a)The active ingredient in household laundry bleaches is sodium hypochlorite (NaClO).2NaOH + Cl2 NaCl + NaClO + H2O
75.0 g NaOH X ---------------------------
1 mol NaOH
40.0 g NaOH
X ------------------------
2 mol NaOH
1 mol NaClO
= 0.938 mol NaClO
71.0 g Cl2
1 mol Cl2X ----------------
1 mol Cl2
1 mol NaClO=0.704 mol NaClO
The calculations show that Cl2 is the limiting reactant. The maximum number of grams of NaClO obtainable from the limiting reactant, that is, the theoretical yield, can now be calculated. It is done using a one-step “moles of A” to “grams of “A” setup.
50.0 g Cl2 X ---------------------------
Theoretical YieldActual Yield and Percent Yield
0.704 mol NaClO X -----------------------75.5 g NaClO
1 mol NaClO= 52.4 g NaClO
Part b
Theoretical YieldActual Yield and Percent Yield
(b) If the actual yield of NaClO for the reaction mixture in part (a) is 43.2 g, what is the percent yield of NaClO for the reaction?
Example 10.13The active ingredient in household laundry bleaches is sodium hypochlorite (NaClO).2NaOH + Cl2 NaCl + NaClO + H2O
actual yieldPercent yield = ----------------------- X 100 = theoretical yield
% yield = --------------------------
52.4 g NaClO
43.2 g NaClOX 100 = 82.4 % NaClO
Actual Yield and Percent Yield
% yield =16.0 g (actual)
52.0 g (theoretical)X 100 = 30.8 %
% yield = ? 16.0 g product 52.0 g th. yield
Theoretical and Percent Yield10.71
% yield =24.79 g (actual)
25.31 g (theoretical)X 100 = 97.95 %
% yield = ? Th. Yield 25.31 and product 24.79 g
Theoretical and Percent Yield10.72
75.0 g Al X ---------------------------
1 mol Al
27.0 g Al
X ------------------------
2 mol Al
1 mol Al2S3
= 1.39 mol Al2S3
32.07 g S
1 mol SX ----------------
3 mol S
1 mol Al2S3= 3.118 mol Al2S3300.0 g S X ---------------------------
Al is the limiting reactant.
Theoretical and Percent Yield10.73
1.39 mol Al2S3 X --------------------------
150. g Al2S3
1 mol Al2S3
= 209 g Al2S3
% yield = ------------------------------
125 g Al2S3 (actual)
209 g Al2S3 (theor)
59.8 %
Theoretical Yield
X 100 =
Theoretical and Percent Yield
Th. Yield = ? 75.0 g Al and 200.0 g O2
Theoretical and Percent Yield
1.39 mol Al2O3 X --------------------------
102 g Al2O3
1 mol Al2O3
= 142 g Al2O3
% yield = ------------------------------
125 g Al2O3 (actual)
142 g Al2O3 (theor)
88.0 %
Theoretical Yield
X 100 =
Theoretical and Actual yield
% yield = ? 2.130 g H2
2.130 g H2 X ---------------------------
1 mol H2
2.016 g H2
X --------------------
1 mol H2
2 mol HCl
= 96.44 % HCl----------------
77.04 g HCl
74.30 g HCl
X 100
% yield =
X ---------------
1 mol HCl
36.46 g HCl
= 77.04 g HCl
Step 2: Find the percent yield of HCl
Step 1: We know from the question that H2 is the limiting reaction. We need to find how many grams of HCl can be produced from 2.130 g H2.
10.75 If 74.30 g of HCl were produced from 2.130 g of H2 and an excess of Cl2 according to reaction H2 + Cl2 2HCl, what was the percent yield of HCl?
----------------Theoretical yield
Actual yield% yield =
X 100
Theoretical and Percent Yield
% yield = ? 28.2 g N2
28.2 g N2 X --------------------------
1 mol N2
28.0 g N2
X -------------------
1 mol Ca3N2
1 mol N2
= 149 g Ca3N2
% Yield =
X -----------------
1 mol Ca3N2
148 g Ca3N2
X 100 =
149 g Ca3N2
115.7 g Ca3N2
--------------------- 77.7%
Theoretical and Actual yield
Actual yield = ? 55.0 g Al
55.0 g Al X --------------------------
1 mol Al
27.0 g Al
X -------------------
1 mol Al2S3
2 mol Al
= 153 g Al2S3
153 g Al2S3 (theor)
X -----------------
1 mol Al2S3
150. g Al2S3
X 100 =
100 Al2S3 (theor)
85.6 g Al2S3 (actual)
--------------------- 131 g Al2S3
Theoretical yield of Al2S3
Actual yield = theoretical yield X % yield
Theoretical and Actual yield 10.77
Actual yield = ? 75.0 g Ag
75.0 g Ag X --------------------------
1 mol Ag
107.9 g Ag
X -------------------
1 mol Ag2S
2 mol Ag
= 86.2 g Ag2S
86.2 g Ag2S (theor)
X -----------------
1 mol Ag2S
248 g Ag2S
X 100 =100 Ag2S (theor)
72.9 g Ag2S (actual)
--------------------- 62.8 g Ag2S
Theoretical yield of Ag2S
Actual yield = theoretical yield X % yield
Theoretical and Actual yield 10.78
g of product = ? 35.0 g CO and O2
35.0 g CO X ---------------------------
1 mol CO
28.0 g CO
X ------------------------
2 mol CO
2 mol CO2
= 1.25 mol CO2
32.0 g O2
1 mol O2X ----------------
1 mol O2
2 mol CO2=2.19 mol CO235.0 g O2 X ---------------------------
The CO is the limiting reactant.
Theoretical and Actual yield 10.79
1.25 mol CO2 X ---------------------------
44.0 g CO2
1 mol CO2
= 55.0 g CO2
55.0 g CO2 (theor) X --------------------
44.0 g CO2 (actual)
100 g CO2 (theor)
= 31.8 g CO2
Theoretical Yield
Actual yield
Theoretical and Actual yield
g of product = ? 2.25 of C and O2
Theoretical and Actual yield 10.80
Simultaneous Reactions
A X ---------------------------
C
B
X ------------------------
D
E
= F
B
CX ----------------
D
E= FA X ---------------------------