modern chemistry chapter 9 stoichiometry. composition stoichiometry deals with the mass...
TRANSCRIPT
Modern Chemistry Chapter 9Modern Chemistry Chapter 9StoichiometryStoichiometry
Modern Chemistry Chapter 9Modern Chemistry Chapter 9StoichiometryStoichiometry
composition composition stoichiometrystoichiometry deals with deals with the mass relationships of the mass relationships of elements in compounds.elements in compounds.
reaction stoichiometryreaction stoichiometry involves the mass involves the mass relationships between relationships between reactants and products in a reactants and products in a chemical reaction.chemical reaction.
Types of Stoichiometry ProblemsTypes of Stoichiometry Problems
1)1) mole to molemole to mole (Both the given (Both the given and the unknown quantities are and the unknown quantities are amounts in moles.)amounts in moles.)
2)2) mole to massmole to mass ( (The given The given amount is in moles and the amount is in moles and the unknown amount is in grams.)unknown amount is in grams.)
3)3) mass to molemass to mole (The given (The given amount is in grams and the amount is in grams and the unknown amount is in moles.)unknown amount is in moles.)
4)4) mass to mass mass to mass (Both the given (Both the given and the unknown amount is in and the unknown amount is in grams.)grams.)
Mole Ratio & Molar MassMole Ratio & Molar Mass
mole ratiomole ratio-- A A conversion factor that conversion factor that relates the amounts in relates the amounts in moles of any two moles of any two substances involved in substances involved in a chemical reaction.a chemical reaction.
Found by using the Found by using the coefficients in the coefficients in the balanced formula balanced formula equation of the equation of the reaction.reaction.
molar mass-molar mass- Equal Equal to the mass in grams to the mass in grams of one mole of an of one mole of an element or a element or a compound.compound.
Found by adding the Found by adding the individual element individual element atomic masses from atomic masses from the formula of the the formula of the compound.compound.
Section Review ProblemsSection Review Problems
Do section review Do section review problems problems
#1 through #4 on #1 through #4 on page 301 of the page 301 of the textbook.textbook.
Section Review page 301Section Review page 301
1- The branch of chemistry that deals with mass 1- The branch of chemistry that deals with mass relationships in compounds and in chemical reactions.relationships in compounds and in chemical reactions.
2 HgO 2 HgO 2 Hg + O 2 Hg + O22
2-2- a) a) 2 mol HgO2 mol HgO & & 2 mol HgO2 mol HgO
2 mol Hg2 mol Hg 1 mol O1 mol O22
2 mol Hg2 mol Hg & & 2 mol Hg2 mol Hg
2 mol HgO2 mol HgO 1 mol O1 mol O22
1 mol O1 mol O22 & & 1 mol O1 mol O22
2 mol HgO2 mol HgO 2 mol Hg2 mol Hg
Section Review page 301Section Review page 301
3- It is used to convert moles of one 3- It is used to convert moles of one substance into moles of another substance into moles of another substance.substance.
4- The formula equation MUST be 4- The formula equation MUST be BALANCED so mole ratios can be BALANCED so mole ratios can be determined.determined.
Using Conversion FactorsUsing Conversion Factors
Amount Amount coefficient coefficient amount of amount of
of of givengiven x x of unknownof unknown = = unknownunknown
substancesubstance coefficient coefficient substancesubstance
in molesin moles of known of known in molesin moles
# moles # moles x mole ratio = #moles ofx mole ratio = #moles of
givengiven unknown unknown
Do practice problems #1 & #2 on page 306 of text.Do practice problems #1 & #2 on page 306 of text.
Page 306 #1Page 306 #1
3 H3 H22 + + N N22 2 NH 2 NH33
6 mol H6 mol H22 x x 2 mol NH 2 mol NH33 = 4mol NH = 4mol NH33
3 mol H3 mol H22
Page 306 #2Page 306 #2
2 KClO2 KClO33 2 KCl + 3 O 2 KCl + 3 O22
15 mol O15 mol O22 x x 2 mol KClO2 mol KClO33 = 10 mol KClO= 10 mol KClO33
3 mol O3 mol O22
Chapter 9 quiz #1- mole to mole calculationsChapter 9 quiz #1- mole to mole calculations
6 NaBr + Mg6 NaBr + Mg33(PO(PO44))22 2 Na 2 Na33POPO44 + 3 MgBr + 3 MgBr22
Use the above balanced formula equation to solve the following:Use the above balanced formula equation to solve the following:
1- 7.0 moles of NaBr will produce ? moles Na1- 7.0 moles of NaBr will produce ? moles Na33POPO44
2- 3.0 moles of Mg2- 3.0 moles of Mg33(PO(PO44))22 will yield ? moles MgBr will yield ? moles MgBr22
3- 0.5 moles of NaBr will react with ? moles Mg3- 0.5 moles of NaBr will react with ? moles Mg33(PO(PO44))22
4- 2.5 moles of NaBr will yield ? moles Na4- 2.5 moles of NaBr will yield ? moles Na33POPO44
5- 2.5 moles NaBr will produce ? moles MgBr5- 2.5 moles NaBr will produce ? moles MgBr22
6 NaBr + Mg6 NaBr + Mg33(PO(PO44))22 2 Na 2 Na33POPO44 + 3 MgBr + 3 MgBr22
1-1- 7 mol NaBr x 7 mol NaBr x 2 mol Na2 mol Na33POPO44 = 2.3 mol Na = 2.3 mol Na33POPO44
6 mol NaBr 6 mol NaBr
2-2- 3 mol Mg3 mol Mg33(PO(PO44))2 2 x x 3.0 mol MgBr3.0 mol MgBr2 2 = 9 mol MgBr= 9 mol MgBr22
1 mol Mg1 mol Mg33(PO(PO44))22
3-3- 0.5 mol NaBr x 0.5 mol NaBr x 1 mol Mg1 mol Mg33(PO(PO44))22 = 0.083 mol = 0.083 mol
6 mol NaBr6 mol NaBr
4-4- 2.5 mol NaBr x 2.5 mol NaBr x 2 mol Na2 mol Na33POPO44 = 0.83 mol = 0.83 mol NaNa33POPO44
6 mol NaBr6 mol NaBr
5-5- 2.5 mol NaBr x 2.5 mol NaBr x 3 mol MgBr3 mol MgBr22 = 1.25 mol MgBr = 1.25 mol MgBr22
6 mol NaBr6 mol NaBr
Using Conversion FactorsUsing Conversion Factors
amount of amount of mass in mass in
givengiven x x moles unknownmoles unknown x molar = grams of x molar = grams of
substancesubstance moles known moles known mass of mass of unknownunknown
in molesin moles unknown unknown substance substance
# moles x mole x molar # moles x mole x molar = mass of unknown = mass of unknown
givengiven ratioratio mass mass (in grams)(in grams)
unknownunknown
Do practice problems #1 & #2 on page 308 of the textbook.Do practice problems #1 & #2 on page 308 of the textbook.
Page 308 #1 & 2Page 308 #1 & 2
2 Mg + O2 Mg + O22 2 MgO 2 MgO
2.00 mol Mg x 2.00 mol Mg x 2 mol MgO2 mol MgO x x 40.3 g MgO40.3 g MgO = 80.6 g MgO = 80.6 g MgO
2 mol Mg2 mol Mg1 mol MgO1 mol MgO
6 CO6 CO22 + 6 H + 6 H22O O C C66HH1212OO66 + 6 O + 6 O22
10 mol CO10 mol CO22 x x 1 mol C1 mol C66HH1212OO66 x x 180 g C180 g C66HH1212OO66 = 300 g C = 300 g C66HH1212OO66
6 mol CO6 mol CO22 1 mol C 1 mol C66HH1212OO66
Chapter 9 Quiz #2- mole-mass problemsChapter 9 Quiz #2- mole-mass problems
3 MgF3 MgF22 + Al + Al22OO33 3 MgO + 2 AlF 3 MgO + 2 AlF33
Use the above balanced formula equation to answer the Use the above balanced formula equation to answer the following questions.following questions.
1- 2.0 mol MgF1- 2.0 mol MgF22 will yield ? grams of MgO will yield ? grams of MgO
2- 4.0 mol of Al2- 4.0 mol of Al22OO33 ? grams of AlF ? grams of AlF33
3- If 6.0 mol of MgO is produced, ? grams of AlF3- If 6.0 mol of MgO is produced, ? grams of AlF33
4- 0.6 mol MgF4- 0.6 mol MgF22 ? grams of AlF ? grams of AlF33
5- 2.75 mol Al5- 2.75 mol Al22OO33 ? grams of MgO ? grams of MgO
3 MgF3 MgF22 + Al + Al22OO33 3 MgO + 2 AlF 3 MgO + 2 AlF33
1- 2.0 mol MgF1- 2.0 mol MgF22 x x 3 mol MgO3 mol MgO x x 40.3 g MgO40.3 g MgO = 80.6 g = 80.6 g
3 mol MgF3 mol MgF22 1 mol MgO 1 mol MgOMgOMgO
2- 4.0 mol Al2- 4.0 mol Al22OO33 x x 2 mol AlF2 mol AlF33 x x 84.0 g AlF84.0 g AlF33 = 672.0g= 672.0g 1 mol Al1 mol Al22OO33 1 mol AlF 1 mol AlF33 AlF AlF33
3- 6.0 mol MgO x 3- 6.0 mol MgO x 2 mol AlF2 mol AlF33 x x 84.0 g AlF84.0 g AlF33 = 336 g AlF = 336 g AlF33
3 mol MgO 1 mol AlF3 mol MgO 1 mol AlF33
4- 0.6 mol MgF4- 0.6 mol MgF22 x x 2 mol AlF2 mol AlF33 x x 84.0 g AlF84.0 g AlF3 3 = 33.6 g AlF = 33.6 g AlF33
3 mol MgF3 mol MgF22 1 mol AlF 1 mol AlF33
5- 2.75 mol Al5- 2.75 mol Al22OO33 x x 3 mol MgO3 mol MgO x x 40.3 g MgO40.3 g MgO = 332gMgO = 332gMgO
1 mol Al1 mol Al22OO33 1 mol MgO 1 mol MgO
Using Conversion FactorsUsing Conversion Factors
mass (g)mass (g) x x 1 mol 1 mol givengiven x x mol mol unknownunknown = moles of = moles of
of of givengiven molar mass molar mass mol mol given unknowngiven unknown
substancesubstance of of givengiven substancesubstance
grams x grams x 11 x mole ratio = moles unknown x mole ratio = moles unknown
molar massmolar mass
Do practice problems #1 & #2 on page 309 of the textbook.Do practice problems #1 & #2 on page 309 of the textbook.
Practice problems page 309Practice problems page 309
2 HgO 2 HgO 2 Hg + O 2 Hg + O22
125 g O125 g O22 x x 1 mol O1 mol O22 x x 2 mol HgO2 mol HgO = 7.81 mol HgO = 7.81 mol HgO 32 g O32 g O22 1 mol O 1 mol O22
125 g O125 g O22 x x 1 mol O1 mol O22 x x 2 mol Hg2 mol Hg = 7.81 mol Hg = 7.81 mol Hg 32 g O32 g O22 1 mol O 1 mol O22
chapter 9 quiz #3- mass-mole problemschapter 9 quiz #3- mass-mole problems
NaNa22O + CaFO + CaF22 2 NaF + CaO 2 NaF + CaO
Use the above equation to solve the problems.Use the above equation to solve the problems.
1- 156.1 grams of CaF1- 156.1 grams of CaF22 ? mol CaO ? mol CaO
2- 186 g Na2- 186 g Na22O O ? mol NaF ? mol NaF
3- 31 g Na3- 31 g Na22O O ? mol CaO ? mol CaO
4- 31 g Na4- 31 g Na22O O ? mol NaF ? mol NaF
5- A yield of 84 g NaF 5- A yield of 84 g NaF ? mol CaO ? mol CaO
NaNa22O + CaFO + CaF22 2 NaF + CaO 2 NaF + CaO
1- 156.1 g CaF1- 156.1 g CaF22 x x 1 mol CaF1 mol CaF22 x x 1 mol CaO1 mol CaO = 2.00 mol CaO = 2.00 mol CaO 78.1 g CaF78.1 g CaF22 1 mol CaF 1 mol CaF22
2- 186.0 g Na2- 186.0 g Na22O x O x 1 mol Na1 mol Na22OO x x 2 mol NaF2 mol NaF = 6.0 mol NaF = 6.0 mol NaF
62 g Na62 g Na22OO 1 mol Na 1 mol Na22O O
3- 31.0 g Na3- 31.0 g Na22O x O x 1 mol Na1 mol Na22OO x x 1 mol CaO1 mol CaO = 0.5 mol CaO = 0.5 mol CaO
62.0 g Na62.0 g Na22O 1 mol NaO 1 mol Na22OO
4- 31.0 g Na4- 31.0 g Na22O x O x 1 mol Na1 mol Na22OO x x 2 mol NaF 2 mol NaF = 1.0 mol NaF = 1.0 mol NaF
62.0 g Na62.0 g Na22O 1 mol NaO 1 mol Na22OO
5- 84.0 g NaF x 5- 84.0 g NaF x 1 mol NaF1 mol NaF x x 1 mol CaO1 mol CaO = 1.0 mol CaO = 1.0 mol CaO
42.0 g NaF 2 mol NaF42.0 g NaF 2 mol NaF
Chemistry Chapter 9- Stoichiometry Practice ProblemsChemistry Chapter 9- Stoichiometry Practice Problems
2 NaF + CaO 2 NaF + CaO Na Na22O + CaFO + CaF22
1-1- 4.5 moles of NaF will produce -?- moles 4.5 moles of NaF will produce -?- moles of of NaNa22O ?O ?
4.5 mol NaF x 1 mol Na4.5 mol NaF x 1 mol Na22O/2 mol NaF = O/2 mol NaF = 2.25 2.25 moles Namoles Na22OO
2-2- 3.2 moles of CaO will produce -?- 3.2 moles of CaO will produce -?- grams of grams of CaFCaF22 ? ?
3.2 mol CaO x 1 mol CaF3.2 mol CaO x 1 mol CaF22/1 mol CaO /1 mol CaO x x 78.1 g CaF78.1 g CaF22/mol CaF/mol CaF22 = 249.9 g CaF = 249.9 g CaF22
2 NaF + CaO 2 NaF + CaO Na Na22O + CaFO + CaF22
3-3- 168.0 grams of NaF will produce -?- 168.0 grams of NaF will produce -?- moles of Namoles of Na22O ?O ?
168.0 /42.0 x 168.0 /42.0 x 1 /2 = 2.0 mol Na1 /2 = 2.0 mol Na22OO
4-4- 112.2 grams of CaO will produce -?- 112.2 grams of CaO will produce -?- moles of CaFmoles of CaF22 ? ?
112.2/56.1 x 1/1 = 2.0 mol CaF112.2/56.1 x 1/1 = 2.0 mol CaF22
5-5- Calculate the molar mass of each of the reactants & Calculate the molar mass of each of the reactants & products of the above balanced formula equation. Use the products of the above balanced formula equation. Use the molar masses in the following problems. molar masses in the following problems.
a-a- AlN = AlN =
(1 x 27.0) + (1 x 14.0) = 41.0 g/mol(1 x 27.0) + (1 x 14.0) = 41.0 g/mol
b-b- NaNa22O =O =
(2 x 23.0) + (1 x 16.0) = 62.0 g/mol(2 x 23.0) + (1 x 16.0) = 62.0 g/mol
c-c- AlAl22OO33 = =
(2 x 27.0) + (3 x 16.0) = 102.0 g/mol(2 x 27.0) + (3 x 16.0) = 102.0 g/mol
d-d- NaNa33N =N =
(3 x 23.0) + (1 x 14.0) = 83.0 g/mol(3 x 23.0) + (1 x 14.0) = 83.0 g/mol
2 AlN + 3 Na2 AlN + 3 Na22O O Al Al22OO33 + 2 Na + 2 Na33NN
6-6- 82.0 grams of AlN will produce -?- 82.0 grams of AlN will produce -?- moles of Almoles of Al22OO33 ? ?
82.0/41.0 x 1/2 = 1.0 mol Al82.0/41.0 x 1/2 = 1.0 mol Al22OO33
7-7- 164.0 grams AlN will produce -?- moles 164.0 grams AlN will produce -?- moles of Naof Na33N ?N ?
164.0/41.0 x 2/2 = 4.0 mol Na164.0/41.0 x 2/2 = 4.0 mol Na33NN
2 AlN + 3 Na2 AlN + 3 Na22O O Al Al22OO33 + 2 Na + 2 Na33NN
8-8- 2.5 moles of Na2.5 moles of Na22O will produce -?- O will produce -?- grams of grams of NaNa33N ?N ?
2.5 x 2/3 x 83.0 = 138.3 g Na2.5 x 2/3 x 83.0 = 138.3 g Na33NN
9-9- 0.75 moles of Na0.75 moles of Na22O will produce -?- O will produce -?- moles moles of Alof Al22OO33 ? ?
0.75 x 1/3 = 0.25 mol Al0.75 x 1/3 = 0.25 mol Al22OO33
10-10- 11.0 moles of Na11.0 moles of Na22O will produce -?- O will produce -?- grams grams of Naof Na33N ?N ?
11.0 x 2/3 x 83.0 = 608.7 g Na11.0 x 2/3 x 83.0 = 608.7 g Na33NN
2 H2 H22 + O + O22 2 H 2 H22OO
11-11- 8.0 grams of H8.0 grams of H22 will react with -?- moles will react with -?- moles of Oof O22 ? ?
8.0/2.0 x 1/2 = 2.0 mol O8.0/2.0 x 1/2 = 2.0 mol O22
12-12- 64.0 grams of O64.0 grams of O22 will produce -?- moles will produce -?- moles of Hof H22O ?O ?
64.0/32.0 x 2/1 = 4.0 mol H64.0/32.0 x 2/1 = 4.0 mol H22OO
13-13- 0.25 moles of H0.25 moles of H22 will produce -?- moles of will produce -?- moles of HH22O ?O ?
0.25 x 2/2 = 0.25 mol H0.25 x 2/2 = 0.25 mol H22OO
2 H2 H22 + O + O22 2 H 2 H22OO
14-14- 1.5 moles of H1.5 moles of H22 will produce -?- grams will produce -?- grams of Hof H22O ?O ?
1.5 x 2/2 x 18.0 = 27.0 g H1.5 x 2/2 x 18.0 = 27.0 g H22OO
15-15- 14 moles of O14 moles of O22 will produce -?- moles will produce -?- moles of of HH22O ?O ?
14 x 2/1 = 28.0 mol H14 x 2/1 = 28.0 mol H22OO
Using Conversion FactorsUsing Conversion Factors
mass (g) x mass (g) x 1 mol 1 mol givengiven x x molmol unknown unknown x x molar mass molar mass unknownunknown = mass = mass
of of givengiven molar mass mol molar mass mol givengiven 1 mol 1 mol unknownunknown of of un-un-
substance of substance of given knowngiven known
grams x grams x 11 x mole ratio x molar mass = mass of x mole ratio x molar mass = mass of unknownunknown
given given molar mass molar mass
Do Do practice problemspractice problems #1, #2, & #3 on page 311 of textbook. #1, #2, & #3 on page 311 of textbook.
Do Do section review problemssection review problems #1 - #5 on page 311 of textbook. #1 - #5 on page 311 of textbook.
Practice problems page 311Practice problems page 311
NHNH44NONO33 N N22O + 2 HO + 2 H22OO
32 g N32 g N22O x O x 1 mol N1 mol N22O O x x 1 mol NH1 mol NH44NONO33 x x 80 g NH80 g NH44NONO33 = 60 g NH = 60 g NH44NONO33
44 g N44 g N22O 1 mol NO 1 mol N22O 1 mol NHO 1 mol NH44NONO33
Chapter 9 quiz #4- mass-mass problemsChapter 9 quiz #4- mass-mass problems
NaNa22O + CaFO + CaF22 2 NaF + CaO 2 NaF + CaO
1- 1- 124 g Na124 g Na22O O ? grams NaF ? grams NaF
2- 2- 124 g Na124 g Na22O O ? grams CaO ? grams CaO
3-3- 234.3 g CaF234.3 g CaF22 ? g NaF ? g NaF
4-4- 234.3 g CaF234.3 g CaF22 ? g CaO ? g CaO
5-5- 84.0 g NaF 84.0 g NaF ? g CaO ? g CaO
NaNa22O + CaFO + CaF22 2 NaF + CaO 2 NaF + CaO
1- 124 g Na1- 124 g Na22O x O x 1 mol Na1 mol Na22OO x x 2 mol NaF2 mol NaF x x 42.0 g NaF42.0 g NaF = 168 g = 168 g
62 g Na62 g Na22O 1 mol NaO 1 mol Na22O 1 mol NaFO 1 mol NaF
2- 124 g Na2- 124 g Na22O x O x 1 mol Na1 mol Na22OO x x 1 mol CaO1 mol CaO x x 56.1 g CaO56.1 g CaO = 112.2 g = 112.2 g
62 g Na62 g Na22O 1 mol NaO 1 mol Na22O 1 mol CaOO 1 mol CaO
3- 234.3 g CaF3- 234.3 g CaF22 x x 1 mol CaF1 mol CaF22 x x 2 mol NaF2 mol NaF x x 42.0 g NaF42.0 g NaF = 252 g = 252 g
78.1 gCaF78.1 gCaF22 1 mol CaF 1 mol CaF22 1 mol NaF 1 mol NaF
4- 234.3 g CaF4- 234.3 g CaF22 x x 1 mol CaF1 mol CaF22 x 1 x 1 mol CaO mol CaO x x 56.1 g CaO56.1 g CaO = 168.3 g = 168.3 g
78.1 gCaF78.1 gCaF22 1 mol CaF 1 mol CaF22 1 mol CaO 1 mol CaO
5- 84.0 g NaF x 5- 84.0 g NaF x 1 mol NaF1 mol NaF x x 1 mol CaO1 mol CaO x x 56.1 g CaO56.1 g CaO = 56.1 g = 56.1 g
42.0 g NaF 2 mol NaF 1 mol CaO42.0 g NaF 2 mol NaF 1 mol CaO
The “MOLE HILL”The “MOLE HILL”
x mole ratio x mole ratio
# moles known# moles known # moles # moles unknownunknown
÷ molar mass÷ molar mass x molar massx molar massof knownof known of unknownof unknown
mass of knownmass of known mass of mass of unknownunknown
Stoichiometry Practice ProblemsStoichiometry Practice Problems
2 H2 H22 + O + O22 2 H 2 H22OO
1)1) 2.5 mol H2.5 mol H22 x x 1 mol O1 mol O22 = 1.25 mol O = 1.25 mol O22
2 mol H2 mol H22
2)2) 2.5 mol H2.5 mol H22 x x 2 mol H2 mol H22O O = 2.5 mol H = 2.5 mol H22OO
2 mol H2 mol H22
3)3) 2.5 mol H2.5 mol H22 x x 1 mol O1 mol O22 x x 32 g O32 g O22 = 40 g O = 40 g O22
2 mol H2 mol H22 1 mol O 1 mol O22
4)4) 2.5 mol H2.5 mol H22 x x 2 mol H2 mol H22OO x x 18 g H18 g H22O O = 45 g H = 45 g H22OO
2 mol H2 mol H22 1 mol H 1 mol H22OO
Stoichiometry Practice ProblemsStoichiometry Practice Problems
2 H2 H22 + O + O22 2 H 2 H22OO
5)5) 16 g H16 g H22 x x 1 mol H1 mol H22 x x 1 mol O1 mol O22 = 4.0 mol O = 4.0 mol O22
2 g H2 g H22 2 mol H 2 mol H22
6)6) 16 g H16 g H22 x x 1 mol H1 mol H22 x x 2 mol H2 mol H22OO = 8.0 mol H = 8.0 mol H22OO 2.0 g H2.0 g H22 2 mol H 2 mol H22
7)7) 16 g H 16 g H22 x x 1 mol H1 mol H22 x x 1 mol O1 mol O22 x x 32 g O32 g O22 = 128 g O= 128 g O22
2.0 g H2.0 g H22 2 mol H 2 mol H22 1 mol O1 mol O22
8)8) 16 g H16 g H22 x x 1 mol H1 mol H22 x x 2 mol H2 mol H22OO x x 18 g H18 g H22O O = 144 g H= 144 g H22OO
2.0 g H2.0 g H22 2 mol H 2 mol H22 1 mol H 1 mol H22OO
Limiting Reactant & Percentage YieldLimiting Reactant & Percentage Yield
limiting reactant limiting reactant is the is the reactant that limits the reactant that limits the amount of the other amount of the other reactant that can reactant that can combine and the amount combine and the amount of product that can be of product that can be formed in a chemical formed in a chemical reaction.reaction.
excess reactantexcess reactant is the is the substance that is NOT substance that is NOT completely used up in a completely used up in a chemical reaction.chemical reaction.
Sample & Practice ProblemsSample & Practice Problems
See sample problem F on page 313 of textbook.See sample problem F on page 313 of textbook.
Do practice problems #1 on page 313.Do practice problems #1 on page 313.
See sample problem G on pages 314-315.See sample problem G on pages 314-315.
Do practice problems #1 & #2 on page 315.Do practice problems #1 & #2 on page 315.
QuickQuickLABLAB
Do the Do the QuickQuickLAB LAB titled “Limiting titled “Limiting Reactants in a Reactants in a Recipe” on page Recipe” on page 316 of the 316 of the textbook.textbook.
Yes, cooking IS Yes, cooking IS chemistry!chemistry!
Percentage YieldPercentage Yield theoretical yield theoretical yield is the maximum is the maximum
amount of product that can be produced amount of product that can be produced from a given amount of reactantfrom a given amount of reactant
actual yield actual yield of a product is the of a product is the measured amount of a product obtained measured amount of a product obtained from a reactionfrom a reaction
percentage yieldpercentage yield is the ratio of the is the ratio of the actual yield to the theoretical yield actual yield to the theoretical yield multiplied by 100multiplied by 100
percentage yield = percentage yield = actual yieldactual yield x 100 x 100theoretical yieldtheoretical yield
ProblemsProblems
see see sample problemsample problem H on pages 317-318 H on pages 317-318 of textbookof textbook
Do Do practice problemspractice problems #1 & #2 on page #1 & #2 on page 318.318.
Do Do section review problemssection review problems #1 - #4 #1 - #4 on page 318.on page 318.
Do Do critical thinking problemscritical thinking problems #37, #38, #37, #38, #39, & #40 on pages 322 & 323.#39, & #40 on pages 322 & 323.
Chapter 9 test reviewChapter 9 test review 20 multiple choice questions20 multiple choice questions
• definitions of composition & reaction definitions of composition & reaction stoichiometrystoichiometry
• mole ratios: their definition & usemole ratios: their definition & use• SI units of molar massSI units of molar mass• identify mole ratio from balanced identify mole ratio from balanced
formula equationformula equation• 5 mole-mole problems5 mole-mole problems• Definitions & practical applications of Definitions & practical applications of
excess reactant & limiting reactantexcess reactant & limiting reactant• definitions & practical applications of definitions & practical applications of
theoretical yield, actual yield, & % yieldtheoretical yield, actual yield, & % yield
Honors Chemistry Chapter 9 Test ReviewHonors Chemistry Chapter 9 Test Review 25 multiple choice25 multiple choice
Know the definitions of reaction & composition Know the definitions of reaction & composition stoichiometry, mole ratio, and units of molar mass.stoichiometry, mole ratio, and units of molar mass.
Know what mole ratio means and how it is used in Know what mole ratio means and how it is used in stoichiometry.stoichiometry.
Determine mole ratio using balanced formula equation. Determine mole ratio using balanced formula equation. (2)(2)
Perform mole to mole stoichiometric calculations (4).Perform mole to mole stoichiometric calculations (4). Perform mole to mass stoichiometric calculations (1).Perform mole to mass stoichiometric calculations (1). Perform mass to mole stoichiometric calculations (1).Perform mass to mole stoichiometric calculations (1). Perform mass to mass stoichiometric calculations (1).Perform mass to mass stoichiometric calculations (1). Know definitions and applications of limiting and excess Know definitions and applications of limiting and excess
reactants.reactants. Know the definitions & applications of actual yield, Know the definitions & applications of actual yield,
theoretical yield, and percent yield. theoretical yield, and percent yield.
Practice #2Practice #2
2 Na2 Na33POPO44 + 3 CaSO + 3 CaSO44 3 Na 3 Na22SOSO44 + Ca + Ca33(PO(PO44))22
Known = 6.0 moles NaKnown = 6.0 moles Na33POPO44 unknown = ? Moles Na unknown = ? Moles Na22SOSO44
6.0 x 3/2 = 9.0 moles Na6.0 x 3/2 = 9.0 moles Na22SOSO44
Known = 5.4 moles CaSOKnown = 5.4 moles CaSO44 unknown = ? Moles Na unknown = ? Moles Na33POPO44
5.4 x 2/3 = 3.6 moles Na5.4 x 2/3 = 3.6 moles Na33POPO44
Known = 0.6 moles NaKnown = 0.6 moles Na33POPO44 unknown = ? Moles CaSO unknown = ? Moles CaSO44
0.6 x 3/2 = 0.9 moles CaSO0.6 x 3/2 = 0.9 moles CaSO44
Stoichiometry Practice #3Stoichiometry Practice #3
4 Na4 Na33N + 3 ON + 3 O22 6 Na 6 Na22O + 2 NO + 2 N22
Assume you have 12.0 moles of NaAssume you have 12.0 moles of Na33N.N.
1)1) How many moles of OHow many moles of O22 do you need? do you need?
2)2) How many moles of NaHow many moles of Na22O will you get?O will you get?
3)3) How many moles of NHow many moles of N22 will you get? will you get?
Stoichiometry Practice #3Stoichiometry Practice #3
4 Na4 Na33N + 3 ON + 3 O22 6 Na 6 Na22O + 2 NO + 2 N22
Assume you have 12.0 moles of NaAssume you have 12.0 moles of Na33N.N.
How many moles of OHow many moles of O22 do you need? do you need?12 x 3/4 = 9 moles O12 x 3/4 = 9 moles O22
How many moles of NaHow many moles of Na22O will you get?O will you get?
12 x 6/4 = 18 moles Na12 x 6/4 = 18 moles Na22OO
How many moles of NHow many moles of N22 will you get? will you get?12 x 2/4 = 6 moles N12 x 2/4 = 6 moles N22
Practice #3Practice #3
4 Na + O4 Na + O22 2 Na 2 Na22OO
If you have 4 moles of Na, how many grams If you have 4 moles of Na, how many grams of Oof O22 will you need? will you need?
If you have 64 grams of OIf you have 64 grams of O22, how many moles , how many moles of Naof Na22O will you produce?O will you produce?
If you have 46 grams of Na, how many If you have 46 grams of Na, how many grams of Ograms of O22 will you need? will you need?
Practice #3Practice #3
4 Na + O4 Na + O22 2 Na 2 Na22OO
If you have 4 moles of Na, how many grams If you have 4 moles of Na, how many grams of Oof O22 will you need? will you need?
4 x 1/4 x 32 = 32 grams O4 x 1/4 x 32 = 32 grams O22
If you have 64 grams of OIf you have 64 grams of O22, how many moles , how many moles of Naof Na22O will you produce?O will you produce?
64/32 x 2/1 = 4 moles Na64/32 x 2/1 = 4 moles Na22OO
If you have 46 grams of Na, how many If you have 46 grams of Na, how many grams of Ograms of O22 will you need? will you need?
46/23 x 1/4 x 32 = 16 grams O46/23 x 1/4 x 32 = 16 grams O22