Stoker Exercises

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


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


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


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


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


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


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


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


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


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


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


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

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

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

http://www.lispme.de/pse/big/04_Be_1.jpg

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

http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA4/STILLS64/AGNO/AGNONA2S/1.JPG



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)


(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


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 ---------------------------


0.704 mol NaClO X -----------------------75.5 g NaClO

1 mol NaClO= 52.4 g NaClO

Part b


(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


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.


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


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


% 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%


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


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.


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


g of product = ? 2.25 of C and O2


Simultaneous Reactions

A X ---------------------------

C

B

X ------------------------

D

E

= F

B

CX ----------------

D

E= FA X ---------------------------

Stoker Exercises

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