reaction stoichiometry

REACTION STOICHIOMETRY

1792 JEREMIAS RICHTERThe amount of substances produced or consumed in chemical reactions can be quantified

4F-1 (of 14)

INFORMATION FROM CHEMICAL EQUATIONS

2H2 + O2 → 2H2O

2 molecules2 moles

0.84 moles0.028 moles

1 molecule1 moles


2 molecules2 moles


4F-2 (of 14)

The moles that react and form do so in the ratio of the balanced equation

INFORMATION FROM CHEMICAL EQUATIONS

2H2 + O2 → 2H2O

0.60 moles-0.60 moles0.00 moles


0.00 moles+0.60 moles

0.60 moles

4F-3 (of 14)

startingreacting

ending



0.00 moles+0.40 moles

0.40 moles

startingreacting

ending

2H2 + O2 → 2H2O

The moles that react and form do so in the ratio of the balanced equation

MASS CALCULATIONS

Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O.

C3H8O + O2 →

CO2 + H2O3 44½

4F-4 (of 14)

MASS CALCULATIONS

Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O.

C3H8O + O2 →

CO2 + H2O6 85.00 g x g2 mol 9 mol

9 2

2 mol C3H8O = 9 mol O2

5.00 g C3H8O

x 9 mol O2 _________________

2 mol C3H8O

= 12.0 g O2

x 1 mol C3H8O ____________________

60.094 g C3H8O

x 32.00 g O2 ______________

1 mol O2

4F-5 (of 14)

Calculate the mass of carbon dioxide produced from the 5.00 g of propanol.

C3H8O + O2 →

CO2 + H2O6 85.00 g x g2 mol 6 mol

9 2

2 mol C3H8O = 6 mol CO2

5.00 g C3H8O

x 6 mol CO2 _________________

2 mol C3H8O

= 11.0 g CO2

x 1 mol C3H8O ____________________

60.094 g C3H8O

x 44.01 g CO2 ________________

1 mol CO2

4F-6 (of 14)

C3H8O + O2 →

CO2 + H2O6 85.0 g 6.0 g

9 2 12.0 g 11.0 g

4F-7 (of 14)

LIMITING REACTANT CALCULATIONS

LIMITING REACTANT – The reactant that is completely used up in a reaction

4F-8 (of 14)

x

12 pieces

x 1 sandwich _______________

1 piece

= 12 sandwiches

22 slices

x 1 sandwich _______________

2 slices

= 11 sandwiches actual amount produced

limiting reactant

12 pieces 22 slices

4F-9 (of 14)

Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride.

Ba + WF3 →

BaF2 + W 25.0 g 26.0 g3 mol 2 mol

323x g

2 mol

2

4F-10 (of 14)

25.0 g Ba

x 2 mol W ____________

3 mol Ba

= 22.3 g W

x 1 mol Ba _______________

137.3 g Ba

x 183.8 g W ______________

1 mol W

Ba + WF3 → 25.0 g 26.0 g3 mol 2 mol

23 BaF2 + W 3x g

2 mol

2


4F-11 (of 14)

25.0 g Ba

x 2 mol W ____________

3 mol Ba

= 22.3 g W

x 1 mol Ba _______________

137.3 g Ba

x 183.8 g W _______________

1 mol W

Ba + WF3 → 25.0 g 26.0 g3 mol 2 mol

23 BaF2 + W 3x g

2 mol

2


26.0 g WF3

x 2 mol W _____________

2 mol WF3

= 19.8 g W

x 1 mol WF3 _________________

240.8 g WF3

x 183.8 g W _______________

1 mol W

WF3 is the limiting reactant19.8 g W are produced

4F-12 (of 14)

Determine the percentage of magnesium and silver in an alloy of the two metals.A 6.50 gram sample of the alloy reacts with 14.5 grams of hydrogen chloride.

Mg + HCl →

MgCl2 + H2

x g 14.5 g1 mol 2 mol

2

1 mol Mg = 2 mol HCl

14.5 g HCl

x 1 mol Mg ______________

2 mol HCl

= 4.834 g Mg

x 1 mol HCl ________________

36.458 g HCl

x 24.31 g Mg ______________

1 mol Mg

4F-13 (of 14)

4.834 g Mg x 100 _______________

6.50 g alloy

= 74.4% Mg 100% - 74.4% = 25.6% Ag

A 1.17 gram sample of the unknown is dissolved in water, treated with lead (II) ions, and 1.42 grams of precipitate are collected.

Pb2+ + I- →

PbI2 x g

2 mol

21.42 g1 mol

1.42 g PbI2

x 2 mol I-

_____________

1 mol PbI2

= 0.7818 g I-

x 1 mol PbI2 _________________

461.0 g PbI2

x 126.9 g I-

_____________

1 mol I-

0.7818 g I-

___________________

1.17 g sample

= 66.8% I- in the sample

x 100

4F-14 (of 14)

Determine the percentage by mass of iodide in a solid unknown.

MOLES FROM SOLUTION DATA

Find the moles of potassium carbonate contained in 275 mL of a 0.300 M potassium carbonate solution.

M = n ___

V

MV = n

x 0.275 L solution = 0.0825 mol K2CO3

0.300 mol K2CO3______________________

L solution

4G-1 (of 12)

Find the moles of each ion in the 0.300 M potassium carbonate solution.

0.0825 mol K2CO3 = 0.165 mol K+

x 2

0.0825 mol K2CO3 = 0.0825 mol CO32-

x 1

4G-2 (of 12)

10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4.

(a) Find the moles of each ion in the solution.

x 0.0100 L solution = 0.004500 mol BaCl2

0.450 mol BaCl2______________________

L solution 0.00450 mol Ba2+ and 0.00900 mol Cl-

x 0.0200 L solution = 0.006000 mol K2SO4

0.300 mol K2SO4______________________

L solution 0.0120 mol K+ and 0.00600 mol SO4

2-

4G-3 (of 12)


(b) Find the moles of each ion after any reaction.

Ba2+ + SO42- → BaSO4

Initial molesReacting molesFinal moles

0.00450 0.00600 0 -0.00450 -0.00450 +0.00450

0 0.00150 0.00450

0 mol Ba2+ 0.00150 mol SO42-

0.00900 mol Cl- 0.0120 mol K+

4G-4 (of 12)


(c) Find the final molarities of each ion in the solution.

= 0 M Ba2+ 0 mol Ba2+ ______________________

0.0300 L solution

= 0.0500 M SO42- 0.00150 mol SO4

2- _______________________

0.0300 L solution = 0.300 M Cl- 0.00900 mol Cl-

_______________________

0.0300 L solution

= 0.400 M K+ 0.0120 mol K+ _______________________

0.0300 L solution

4G-5 (of 12)

20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH.

(a) Find the moles of each ion in the solution.

x 0.0200 L solution = 0.007000 mol HCl

0.350 mol HCl___________________

L solution 0.00700 mol H+ and 0.00700 mol Cl-

x 0.0300 L solution = 0.007500 mol NaOH

0.250 mol NaOH______________________

L solution 0.00750 mol Na+ and 0.00750 mol OH-

4G-6 (of 12)


(b) Find the moles of each ion after any reaction.

H+ + OH- → H2O

Initial molesReacting molesFinal moles

0.00700 0.00750 0 -0.00700 -0.00700 +0.00700

0 0.00050 0.00700

0 mol H+ 0.00050 mol OH-

0.00700 mol Cl- 0.00750 mol Na+

4G-7 (of 12)


(c) Find the final molarities of each ion in the solution.

= 0 M H+ 0 mol H+ ______________________

0.0500 L solution

= 0.010 M OH- 0.00050 mol OH- _______________________

0.0500 L solution

= 0.140 M Cl- 0.00700 mol Cl- _______________________

0.0500 L solution

= 0.150 M Na+ 0.00750 mol Na+ _______________________

0.0500 L solution

4G-8 (of 12)


(d) Find the mass of water produced by the reaction.

= 0.126 g H2O 0.007000 mol H2O

x 18.016 g H2O __________________

1 mol H2O

4G-9 (of 12)

DILUTION CALCULATIONS

When a solution is diluted, only solvent is added , so the moles of soluteare unchanged

mol solute (concentrated) = mol solute (diluted)MCVC = MDVD

4G-10 (of 12)

Calculate the volume of 6.00 M ammonia needed to prepare 250. mL of a 0.100 M ammonia solution.

MCVC = MDVD

VC = MDVD _______

MC

= (0.100 M)(250. mL) ________________________

(6.00 M)

= 4.17 mL

MC =VC =

6.00 M?

MD =VD =

0.100 M250. mL

4G-11 (of 12)

Calculate the volume of water that must be added to 5.00 mL of concentrated hydrochloric acid (12.1 M) to make the acid 3.00 M.

MCVC = MDVD

MCVC = VD_______

MD

= (12.1 M)(5.00 mL) _______________________

(3.00 M)

= 20.2 mL

MC =VC =

12.1 M5.00 mL

MD =VD =

3.00 M?

20.2 mL- 5.00 mL____________

Volume of dilute solutionVolume of concentrated solution

15.2 mL Water that must be added

4G-12 (of 12)

TITRATION – A technique in which one solution is used to analysis another

Buret: a solution of 1 reactant of known concentration

REACTIONS IN SOLUTION

Flask: another reactant of unknown concentration, mass, etc.

STANDARD SOLUTION – A solution of known concentration

4H-1 (of 13)

The mass of sodium bicarbonate in an antacid tablet is to be determined.ACID-BASE INDICATOR – A weak organic acid or base that changes color in acidic or basic solutionsThe tablet is dissolved in water, an acid-base indicator added, and 21.5 mL of a 0.300 M hydrochloric acid solution produces a color change.

NaHCO3 + HCl →

NaCl + H2O + CO2 x g 21.5 mL

0.300 M1 mol 1 mol

x 1 mol NaHCO3 _________________

1 mol HCl

0.300 mol HCl x 0.0215 L solution __________________

L solution

= 0.542 g NaHCO3

x 84.008 g NaHCO3 _______________________

mol NaHCO3

4H-2 (of 13)

x 1 mol O.A.D. ____________________

126.08 g O.A.D.

A sodium hydroxide solution is to be standardized.34.2 mL of the sodium hydroxide solution are required to neutralize a solution made with 0.619 grams of solid H2C2O4

.2H2O (m = 126.08 g/mol).

NaOH + H2C2O4 → .2H2O Na2C2O4 + H2O 34.2 mL

x M0.619 g

2 mol 1 mol

x 2 mol NaOH _________________

1 mol O.A.D.

0.619 g O.A.D.

= 0.287 M NaOH

x 1 _______________________

0.0342 L solution

2 24

_

4H-3 (of 13)

A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL of a 0.115 M MnO4

- solution produces a light purple color. If the V2+ was oxidized to V5+, determine the molarity of the V2+ in the original solution.

4H-4 (of 13)

( ) x 3

( ) x 5

MnO4- (aq) + V2+ (aq) →

MnO4- → Mn2+

+7 -2 +2

V2+ → V5+

+ 4H2O 8H+ +5e- +

+ 3e-

Mn2+ + V5+

+2 +5



4H-5 (of 13)

3MnO4- → 3Mn2+

5V2+ → 5V5+

+ 12H2O 24H+ +15e- +

+ 15e-

15e- + 24H+ + 3MnO4- + 5V2+ →

3Mn2+ + 12H2O + 5V5+ + 15e-

MnO4- (aq) + V2+ (aq) →

+7 -2 +2 +2 +5Mn2+ + V5+



10.0 mL x M

32.7 mL0.115 M

5 mol3 mol

x 5 mol V2+

________________

3 mol MnO4-

0.115 mol MnO4- x 0.0327 L solution

_____________________

L solution

= 0.627 M V2+

x 1 _______________________

0.0100 L solution

4H-6 (of 13)

3MnO4- (aq) + 5V2+(aq) → 3Mn2+ (aq) + 12H2O (l) + 5V5+ (aq) 24H+

(aq) +

molar mass H2X = grams H2X_______________

moles H2X

molar mass H2X = 0.109 grams H2X ______________________

? moles H2X

Calculate the molar mass of a diprotic acid if 0.409 grams of it are neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution.

4H-7 (of 13)

H2X + NaOH →

H(OH) + Na2X 0.409 gx mol

19.50 mL0.287 M

1 mol 2 mol

2

x 1 mol H2X _______________

2 mol NaOH

0.287 mol NaOH x 0.01950 L sol’n _____________________

L sol’n

= 0.002798 mol H2X

2

Calculate the molar mass of a diprotic acid if 0.409 grams of it are neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution.

0.409 g H2X________________________

0.002798 mol H2X

= 146 g/mol

4H-8 (of 13)

The molarity of an aluminum hydroxide solution is to be determined.An acid-base indicator is added to 10.0 mL of the aluminum hydroxide solution, and 12.5 mL of 0.300 M hydrochloric acid produces a color change.

Al(OH)3 + HCl →

AlCl3 + H(OH) 10.0 mL

x M12.5 mL0.300 M

1 mol 3 mol

3

x 1 mol Al(OH)3 _________________

3 mol HCl

0.300 mol HCl x 0.0125 L solution __________________

L solution

= 0.125 M Al(OH)3

x 1 ____________

0.0100 L

3

4H-9 (of 13)

( ) x 8

Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O7

2- solution. Determine the percentage of S2- in cinnabar ore.

4H-10 (of 13)

( ) x 3

K2Cr2O7 (aq) + S2- (aq) →

K+ (aq) + Cr2O7

2- (aq) + S2- (aq) →

Cr2O72- → Cr3+

+1 +6 -2 -2

S2- → S8

2

8

+ 7H2O14H+ +6e- +

+ 16e-

Cr3+ + S8

+3 0



4H-11 (of 13)

K2Cr2O7 (aq) + S2- (aq) →

K+ (aq) + Cr2O7

2- (aq) + S2- (aq) →

8Cr2O72- → 16Cr3+

+1 +6 -2 -2

24S2- → 3S8

+ 56H2O112H+ +48e- +

+ 48e-

48e- + 112H+ + 8Cr2O72- + 24S2- → 16Cr3+ + 56H2O + 3S8 + 48e-

Cr3+ + S8

+3 0



x g20.4 mL0.110 M

x 24 mol S2-

_________________

8 mol Cr2O72-

0.110 mol Cr2O72- x 0.0204 L solution

_______________________

L solution

= 0.2159 g S2-

x 32.07 g S2- _____________

mol S2-

0.2159 g S2- x 100 _______________

1.534 g ore

= 14.1% S2- in the ore

4H-12 (of 13)

8Cr2O72- (aq) + 24S2-(aq) → 16Cr3+ (aq) + 56H2O (l) + 3S8

(s) 112H+ (aq) +

8 mol 24 mol

Calculate the molarity of a sulfuric acid solution if 25.0 mL of it are neutralized by 33.5 mL of a 0.240 M potassium hydroxide solution.

H2SO4 + KOH →

H(OH) + K2SO4 25.0 mL

x M33.5 mL0.240 M

1 mol 2 mol

2

x 1 mol H2SO4 _________________

2 mol KOH

0.240 mol KOH x 0.0335 L solution ___________________

L solution

= 0.161 M H2SO4

x 1 ____________

0.0250 L

2

4H-13 (of 13)

MOLAR MASSES AND STOICHIOMETRIC CONVERSIONS

2 mol C (12.01 g/mol) = 24.02 g6 mol H (1.008 g/mol) = 6.048 g

1 mol O (16.00 g/mol) = 16.00 g46.068 g

Calculate the molar mass of ethanol, C2H5OH

4I-1 (of 8)

46.068 g C2H5OH = 1 mol C2H5OH

x 1 mol C2H5OH ______________________

46.068 g C2H5OH

Calculate the number of ethanol molecules in 25.0 mL of pure ethanol. The density of the ethanol is 0.789 g/mL.

25.0 mL C2H5OH

x 6.022 x 1023 molecules C2H5OH ________________________________________

1 mol C2H5OH

= 2.58 x 1023 molecules C2H5OH

x 0.789 g C2H5OH ____________________

1 mL C2H5OH

0.789 g C2H5OH = 1 mL C2H5OH

4I-2 (of 8)

x 0.789 g C2H5OH ____________________

1 mL C2H5OH

x 1 mol C2H5OH ______________________

46.068 g C2H5OH

Calculate the number of carbon atoms in a 10.0 mL sample of pure ethanol.

10.0 mL C2H5OH

x 2 mol C __________________

1 mol C2H5OH

= 2.06 x 1023 atoms C

x 6.022 x 1023 atoms C ___________________________

1 mol C

4I-3 (of 8)

x 40. mL C2H5OH ____________________

100 mL vodkax 1 mol C2H5OH ______________________

46.068 g C2H5OH

Calculate the number of ethanol molecules in 45.0 mL of 80. proof vodka. The density of the vodka is 0.92 g/mL.

45.0 mL vodka

x 6.022 x 1023 molecules C2H5OH ________________________________________

1 mol C2H5OH

= 1.9 x 1023 molecules C2H5OH

x 0.789 g C2H5OH ____________________

1 mL C2H5OH

80. Proof vodka = 40.% C2H5OH by volume

100 mL vodka = 40. mL C2H5OH

4I-4 (of 8)

x 46.068 g C2H5OH _______________________

mol C2H5OH

Calculate the mass of one ethanol molecule, in grams.

= 7.650 x 10-23 g

x mol C2H5OH ________________________________________

6.022 x 1023 molecules C2H5OH

4I-5 (of 8)

1 molecule C2H5OH

x 1 mol O ____________

16.00 g O

A metal oxide with the formula M2O3 is 29.0% oxygen by mass. Calculate the molar mass of metal M.

29.0 g O x 2 mol M __________

3 mol O

= 58.8 g/mol 71.0 g M_________________

1.208 mol M

4I-6 (of 8)

g Mmol M

Molar Mass of M = = 71.0 g M ? mol M

= 1.208 mol M

1 mol Na (22.99 g/mol) = 22.99 g1 mol N (14.01 g/mol) =

14.01 g3 mol O (16.00 g/mol) =48.00 g

85.00 g

THEORETICAL PERCENT COMPOSITION OF COMPOUNDS BY MASS

NaNO3

Calculate the percent composition by mass of sodium nitrate

22.99 g Na 100___________________

85.00 g NaNO3

% Na = = 27.05 % Na

48.00 g O 100___________________

85.00 g NaNO3

% O = = 56.47 % O

14.01 g N 100___________________

85.00 g NaNO3

% N = = 16.48 % N

4I-7 (of 8)

BaCl2.2H2O

1 mol Ba (137.3 g/mol) = 137.3 g2 mol Cl (35.45 g/mol) = 70.90 g

2 mol H2O (18.016 g/mol) = 36.032 g244.232 g

Calculate the percentage by mass of water in barium chloride dihydrate

36.032 g H2O 100____________________________

244.232 g BaCl2.2H2O

% H2O = = 14.75 % H2O

4I-8 (of 8)

EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS

When a sample of a hydrocarbon is burned, 8.45 g CO2 and 1.73 g H2O are produced. Calculate the empirical formula of the hydrocarbon.

CxHy + O2 →

CO2 + H2O

8.45 g CO2

= 2.306 g C x 12.01 g C ________________

44.01 g CO2

1.73 g H2O

= 0.1936 g H x 2.016 g H ________________

18.016 g H2O

4J-1 (of 9)

x 1 mol C_____________

12.01 g C

= 0.1920 mol C

2.306 g C

x 1 mol H____________

1.008 g H

= 0.1921 mol H

0.1936 g H

0.1920 mol C_________________

0.1920

0.1921 mol H_________________

0.1920

= 1.00 mol C = 1.00 mol H

Empirical formula: CH

EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS

When a sample of a hydrocarbon is burned, 8.45 g CO2 and 1.73 g H2O are produced. Calculate the empirical formula of the hydrocarbon.

4J-2 (of 9)

A borane is any compound composed of boron and hydrogen. When a sample of a borane is burned, 12.89 g B2O3 and 12.27 g H2O are produced. Calculate the empirical formula of the borane.

BxHy + O2 →

B2O3 + H2O

12.89 g B2O3

= 4.0029 g B x 21.62 g B ________________

69.62 g B2O3

12.27 g H2O

= 1.3730 g H x 2.016 g H ________________

18.016 g H2O

4J-3 (of 9)

x 1 mol B_____________

10.81 g B

= 0.37030 mol B

4.0029 g B

x 1 mol H_____________

1.008 g H

= 1.3621 mol H

1.3730 g H

0.37030 mol B___________________

0.37030

1.3621 mol H_________________

0.37030

= 1.000 mol B = 3.678 mol H

Empirical formula: B3H11

A borane is any compound composed of boron and hydrogen. When a sample of a borane is burned, 12.89 g B2O3 and 12.27 g H2O are produced. Calculate the empirical formula of the borane.

4J-4 (of 9)

x 3 x 3

When a 3.84 g sample of a compound containing C, H, and N is combusted, 7.34 g CO2 and 2.51 g H2O are collected. Calculate the empirical formula of the compound.

CxHyNz + O2 → CO2 + H2O + NaOb

7.34 g CO2

= 2.003 g C x 12.01 g C ________________

44.11 g CO2

2.51 g H2O

= 0.2809 g H x 2.016 g H _____+__________

18.016 g H2O

3.84 g CxHyNz

- 2.003 g C- 0.2809 g H

________________________

1.5561 g N

4J-5 (of 9)

When a 3.84 g sample of a compound containing C, H, and N is combusted, 7.34 g CO2 and 2.51 g H2O are collected. Calculate the empirical formula of the compound.

x 1 mol C_____________

12.01 g C

= 0.1668 mol C

2.003 g C

x 1 mol H____________

1.008 g H

= 0.2787 mol H

0.2809 g H

0.1668 mol C_________________

0.1111

0.2787 mol H_________________

0.1111= 1.50 mol C = 2.51 mol H

Empirical formula: C3H5N2

x 1 mol N_____________

14.01 g N

= 0.1111 mol N

1.5561 g N

0.1111 mol N________________

0.1111= 1.00 mol N

4J-6 (of 9)

When a 2.75 g sample of a compound containing C, H, and O is combusted, 5.49 g CO2 and 2.25 g H2O are collected. Calculate the empirical formula of the compound.

CxHyOz + O2 → CO2 + H2O

5.49 g CO2

= 1.498 g C x 12.01 g C ________________

44.11 g CO2

2.25 g H2O

= 0.2518 g H x 2.016 g H _________________

18.016 g H2O

2.75 g CxHyOz

- 1.498 g C- 0.2518 g H

________________________

1.0002 g O

4J-7 (of 9)

When a 2.75 g sample of a compound containing C, H, and O is combusted, 5.49 g CO2 and 2.25 g H2O are collected. Calculate the empirical formula of the compound.

x 1 mol C____________

12.01 g C

= 0.1247 mol C

1.498 g C

x 1 mol H____________

1.008 g H

= 0.2498 mol H

0.2518 g H

0.1247 mol C_________________

0.06251

0.2498 mol H_________________

0.06251= 1.99 mol C = 4.00 mol H

Empirical formula: C2H4O

x 1 mol O_____________

16.00 g O

= 0.06251 mol O

1.0002 g O

0.06251 mol O__________________

0.06251= 1.00 mol O

4J-8 (of 9)

C2H4O

2 mol C (12.01 g/mol) = 24.02 g4 mol H (1.008 g/mol) = 4.032 g1 mol O (16.00 g/mol) = 16.00 g

44.052 g

90 g/mol________________

44.052 g/mol

≈ 2

Molecular formula: C4H8O2

Calculate the molecular formula of the previous compound if it has a molar mass of about 90 g/mol.

4J-9 (of 9)

reaction stoichiometry

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mol c3h8o

g c3h8o x

g o2 x

g of tungsten

g ba x

mol ba

g co2 x

g wf3 x