Mass Relationships in Chemical Reactions Chapter 3 4 - 5 Lectures Dr. Ali Bumajdad.
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Transcript of Mass Relationships in Chemical Reactions Chapter 3 4 - 5 Lectures Dr. Ali Bumajdad.
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Mass Relationships in Chemical Reactions
Chapter 34 - 5 Lectures
Dr. Ali Bumajdad
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Chapter 3 Topics
Dr. Ali Bumajdad
• Average Atomic Masses•The Mole•Molar Mass (M.m.) and Molecular mass (M.w.) •Percent Composition•Calculation Empirical & Molecular Formula•Balancing Chemical Equation•Stoichiometric Calculations & Limiting Reactant•Theoretical yield, actual yield and percentage yield
Stoichiometry
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By definition: 1 atom 12C “weighs” 12 amu
On this scale
1H = 1.008 amu
16O = 16.00 amu
•Atomic mass is the mass of an atom in atomic mass units (amu)
Micro Worldatoms & molecules
AmuAtomic mass
Macro Worldgrams
Molar mass
Atomic mass & Average Atomic Masses
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Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
7.42 x 6.015 + 92.58 x 7.016100
= 6.941 amu
Average atomic mass of lithium:
Dr. Ali Bumajdad
Av. At. mass = (m of Isotope 1 x its abundance) + (m of Isotope 2 x its abundance) + …….100
(0)
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Average atomic mass (6.941)
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Sa Ex 3.1: Cu vaporized in a mass spectrometer and It was found to have two isotopes one of mass 62.93 amu and abundance 69.09% and the other of mass 64.93 amuand abundance 30.91%
Find its average mass.
Av. m = 63.55 amu
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•mole : amount of a substance that contains 6.0221367 x 1023 objects.
•mole : number of carbon atoms in 12 g of 12C1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
number
number
number
•The Mole SI unit of amount of substance
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12.00 grams of 12C contains 6.022 x 1023
1.008 grams of H contains 6.022 x 1023
Mass of 1 mole
6.00 grams of 12C contains 3.011 x 1023
Number of atoms in 1 mole
Mass of 1/2 mole Number of atoms in 1/2 mole
Dr. Ali Bumajdad
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Molar mass is the mass of 1 mole of in gramseggsshoes
marblesatoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass in amu = molar mass in grams/mol
Molar Mass (M.m.) and Molecular mass (M.w.)
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C 12.01 amu mass of 1 atom of C12.01 g mass of 1mol of C
mass of 6.022 x 1023 atoms of C
•From the periodic table
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One Mole of atoms
C S
Cu Fe
Hg
32 g
55.85 g
200.59 g
12.01 g
63.55 g
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Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.
SO2
1S 32.07 amu
2O + 2 x 16.00 amu SO2 64.07 amu
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
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For any compound
Molecular mass in amu = molar mass in grams/mol
M.w. of SO2 = 64.07 amu
M.w. of C8H10N4O2 = 194.20 amu
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One Mole of molecules
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Q) 1 mol of apple contains 6.022 x 1023 apple
Q) 1 mol of CH3OH contains 6.022 x 1023 CH3OH
Q) 1 mol of CH3OH contains 6.022 x 1023 O atoms
Q) 1 mol of CH3OH contains 4 x 6.022 x 1023 H atoms
Dr. Ali Bumajdad
Q) 1/2 mol of CH3OH contains (4 x 6.022 x 1023)/2 H atoms
Q) 1/2 mol of CH3OH contains (6.022 x 1023)/2 CH3OH
N = n × NA
N = no. of dozen × 12
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M.m. = molar mass in g/mol
NA = Avogadro’s number
N = number of objects
n = number of mole
m = mass
M.m. n =
m(1)
NA n =
N(2)
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No. of A atomsNo. of AaBb molecules
No. of AaBb molecules
× a
a
AaBb
× a 6.022×1023
6.022×1023 × a No. of A moles
Q) 1.0 ×103 CH4 molecules contain:
1) How many H atoms.2) How many moles of H atoms.
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Q) How many atoms are in 0.551 g of potassium (K) ?
I need n and NA
Find n using:
n = 0.551 g / 39.10 g mol-1
n = 0.0141 mol
N = (0.0141) (6.022 x 1023) = 8.49 x 1021 K atoms
NA n =
N(2)
Expected Dr. Ali Bumajdad
M.m. n =
m(1)
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Q) How many H atoms are in 72.5 g of C3H8O ?
Answer = 5.82 x 1024 atoms H
Dr. Ali Bumajdad
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NA m of 1 atom in g =
m of 1 mol in g(3)
12 m of 1 apple in g =
m of 1 dozen in g
Because
NA m of 1 atom in g =
M.m.(3)
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Sa Ex. 3.2: Calculate the mass in gram of 6 atoms of Americium (Am)
Method 1: Using Eq. 3
Method 2:
243 g 6.022 x 1023 atoms
X 6 atoms
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Sa Ex. 3.3: Calculate the number of moles and the number of atoms in 10.0 g sample of Al.
Use Eq. 1 and 2
n = 0.371 mol Al
N = 2.23 × 1023 Al atoms
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Sa Ex. 3.4: How many Si atoms in 5.68 mg of silicon computer chip?
N = 1.22 × 1020 Si atoms
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Sa Ex. 3.5: Calculate the number of moles and the mass in gram of a sample of Co containing 5.00 ×1020 atoms
m of 5.00 ×1020 atoms = 4.89 ×10-2g
n = 8.30 ×10-4 mol
Method 1:
Method 2:n = 8.30 ×10-4 mol
m of 5.00 ×1020 atoms = 4.89 ×10-2g
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Sa Ex. 3.6: (1) Calculate the molar mass of C10H6O3. (2) A sample of 1.56 ×10-2g of C10H6O3,
how many moles does this sample represent
(1) M.m.= 174.1g(2) No. of moles = 8.96 × 10-5 mol
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Sa Ex. 3.7: (1) Calculate the molar mass of CaCO3. (2) A sample of 4.86 moles of CaCO3,
what is the mass of the CO32- ions present?
M.m. = 100.09 g/mol
mass of the CO32- = 292 g
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Sa Ex. 3.8: Bees release Isopentyl acetate (C7H14O2) when sting. The amount release is about 1 g (1) How many molecules of C7H14O2 are released (2) How many atoms of Carbon are present (3) How many atoms are present
(1) no. of molecules= 5×1015 molecules(2) no. of C atoms = 4× 1016 C atoms(3) no. of atoms = 1× 1017 atoms
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M.m. n =
m(1)
NA n =
N(2)
We should use Eq.1 and Eq. 2 and …
Number of H atoms = 1.03 ×1024 H atoms
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Mass of Zn = 23.3 gram
M.m. n =
m(1)
We should use Eq.1
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We should use Eq.1 and Eq. 2
NA n =
N(2)
Number of S atoms = 3.06 × 1023 S atoms
M.m. n =
m(1)
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Ligh
t
Ligh
t
Hea
vy
Hea
vy
•Percent Composition
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%C =(12.01 g) x 2
46.07 gx 100% = 52.14%
%H =(1.008 g) x 6
46.07 gx 100% = 13.13%
%O =(16.00 g) x 1
46.07 gx 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
Dr. Ali Bumajdad
A) If I know M.F.
C2H6O
Suppose I have AaBb molecule
(4)% A =
(M.m.A) × a
M.m. AaBb
x 100%
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Suppose I have AaBb molecule
Dr. Ali Bumajdad
B) If I do not know M.F. but I know the masses
(5)% A =
m A
mA +mB
x 100%
CyHx
m of C =24 g m of H = 8g
%C =24 g
32 gx 100% = 75 %
%H =8 g
32 gx 100% = 25 %
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% H = = 3.086%% P = = 31.61%% O = = 65.31%
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1) Find the moles of H and C from mass of H2O and CO2
2) Find mass of O (or any other element) using:
Mass of O = mtotal – (mH+mC)
3) Find the moles of O from mass of O
4) Divide by the smallest mole value
A) Empirical formula from combustion of known amount
•Calculation Empirical & Molecular Formula
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Combust 11.5 g ethanol
13.5 g H2O
0.5 mol C
1.5 mol H
0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
g CO2 mol CO2 mol C
g H2O mol H2O mol H
g O mol O
22.0 g CO2
Dr. Ali Bumajdad
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Q) 0.1156g f compound contains C, H and N only. The H2OAbsorber increase by 0.1676g and the CO2 absorber increase by 0.1638g what is the empirical formal of the compound?
The E.F. is CH5N
Dr. Ali Bumajdad
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C) Molecular Formula from element masses or %mass and M.m.M.F.
M.F. = E.F. ×M.m. E.F.
M.m. M.F.(6)
1) Find the moles of elements (for %mass assume you have 100 g)
2) Divide by the smallest mole value. Now you know E.F.3) Use:
B) Empirical Formula from element masses or %mass
1) Find the moles of elements (for %mass assume you have 100 g)
2) Divide by the smallest mole value
Dr. Ali Bumajdad
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M.F. = N2O4
M.m. = 90.02 g/mol
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Sa. Ex. 3.11: Compounds consists of 71.65% Cl, 24.27%CAnd 4.07% H and its molar mass = 98.96 g/mola) Determine its E.F.b) Determine its M.F.
(Assume that we have 100 grams of compound)
a)E.F. = CH2Clb)M.F. = C2H4Cl2
Dr. Ali Bumajdad
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Sa. Ex 3.12: White powder contain 43.64% P and 56.36% O. The compound molar mass = 283.88 g/mol.
a) What is E.F.b) What is M.F.
E.F. = P2O5
M.F. = P4O10
Dr. Ali Bumajdad
1) Find the moles of elements (for %mass assume you have 100 g)
2) Divide by the smallest mole value
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Balancing Chemical Equation
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
1 moles Mg + 1/2 mole O2 makes 1 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgONOT
2 grams Mg + 1 gram O2 makes 2 g MgO
Reactant Product
Coefficent
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Balancing Chemical EquationsC2H6 + O2 CO2 + H2O
1.Start by balancing those elements that appear in only one reactant and one product.
start with C or H but not O
C2H6 + O2 2CO2 + 3H2O
2. Balance those elements that appear in two or more reactants or products.
C2H6 + O2 2CO2 + 3H2O72
2C2H6 + 7O2 4CO2 + 6H2O
Dr. Ali Bumajdad
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Q) Balance:
1) 1C2H5OH (L) + 3O2 (g) 2CO2 (g) + 3H2O (g)
2) Na2CO3 (s) + 2HCl (aq) 2NaCl (s) + H2O (L) + CO2 (g)
3) 1C6H6 (L) + 7.5 O2 (g) 6 CO2 (g) + 3H2O (g) 2 15 12 6
Sa Ex. 3.14: balance
(NH4)2Cr2O7 (s) Cr2O3 (s) + N2(g) + 4 H2O (g)
Sa Ex. 3.15: balance
2NH3 (g) + 5/2O2 (g) 2NO (g) + 3H2O (g) 4 5 4 6 Dr. Ali Bumajdad
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4Al + 3O2 2Al2O3
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1. Balanced the equation
2. Find moles of one reactant or product
3. Use coefficient ratio to calculate the moles of the required substance
4. Convert moles of required substance into masses
Calculation involving masses
Stoichiometric Calculations & Limiting Reactant
( Coefficient provide moles ratios not exact amount)
Dr. Ali Bumajdad
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Q) How many moles of C atoms are needed to combined with 4.87 mol Cl to form C2Cl6
1) Balanced the equation
2C + 3Cl2 C2Cl6
2) Find moles of one reactant or product
2C + 3Cl2 C2Cl6
2.435 mol?3) Use coefficient ratio to calculate the moles of the required substance
2 mol of C 3 mol Cl2 X 2.435 mol of Cl2
= 1.62 mol CDr. Ali Bumajdad
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Q) 209 g of methanol burns in air according to the equation:
Find mass of water formed.
CH3OH + O2 CO2 + H2O
m H2O =n ×M.m.=13.04×18.016 = 235 g H2O
1) Balanced the equation
2) Find moles of one reactant or product
Dr. Ali Bumajdad
2CH3OH + 3O2 2CO2 + 4H2O
2CH3OH + 3O2 2CO2 + 4H2O
209 g
32.04 g/moln = = 6.52 mol
4) Convert moles of required substance into masses
M.m. n =
m (1)
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of CH3OH 4 mol H2O6.52 mol of CH3OH X = 13.04 mol H2O
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Q) How many grams of Ca must react with 83.0 g of Cl2 to form CaCl2
1) Balanced the equationCa + Cl2 CaCl2
2) Find moles of one reactant or product
3) Use coefficient ratio to calculate the moles of the required substance
83.0 g?
Ca + Cl2 CaCl2
83.0 g
70.9 g/moln = = 1.17 mol
4) Convert moles of required substance into masses
m Ca =n ×M.m.= 1.17×40.08 = 46.9 g CaM.m. n =
m (1)
= 1.17 mol Ca
1 mol of Ca 1 mol Cl2 X 1.17 mol of Cl2
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Sa. Ex. 3.16: LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L)What is the mass of CO2 require to react with 1.00 Kg of LiOH?
1) Balanced the equation
2) Find moles of one reactant or product
3) Use coefficient ratio to calculate the moles of the required substance
4) Convert moles of required substance into masses
m CO2 =n ×M.m.= 20.88×44.0 = 920 g CO2
M.m. n =
m (1)
2LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L)
1.00 × 103 g ?
1.00 × 103 g
23.95 g/moln = = 41.75 mol (expected)
2LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L)
2 mol of LiOH 1 mol CO2
41.75 mol of LiOH X = 20.88 mol CO2
Dr. Ali Bumajdad
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6 green used up
6 red left over
Limiting Reactant
Dr. Ali Bumajdad
(the reactant the completely consumed in a chemical reaction)
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2 mol 1mol 2mol
1 mol 1/2mol 1mol
2.1 mol 1mol 2mol
1.9 mol 1mol 1.9mollimiting
limiting
Limiting reactant determine the amount of product
2H2 + O2 2H2Oe.g.
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• Why limiting reactant is important?
Because in the stoichiometric calculation I should only use the coefficient of the limiting reactant
• How do I know that a reaction contains a limiting reactant ?
If I've been given the masses or number of moles of two reactant then I might have limiting reactant
Dr. Ali Bumajdad
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1. Balanced the equation
2. Find moles of the two reactants
3. Identify the limiting reactant (How?)
4. Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance
5. Convert moles of required substance into masses
Calculation involving a limiting reactant
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Q) In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.1) Balanced the equation
2) Find moles of the two reactants
2Al + Fe2O3 Al2O3 + 2Fe
Dr. Ali Bumajdad
3) Identify the limiting reactant (How?)
4.60
2= 2.3
3.76
1= 3.76
limiting
2Al + Fe2O3 Al2O3 + 2Fe124g ?
124 g
26.98 g/moln = = 4.60 mol
601g
601 g
159.69 g/moln = = 3.76 mol
limiting
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Dr. Ali Bumajdad
3) Use the coefficient ratio between the limiting reactant
and the required substance to find out the number of mole
of the required substance
4) Convert moles of required substance into masses
m Al2O3 =n ×M.m.= 2.30×101.96 = 235 g Al2O3M.m. n =
m (1)
= 2.30 mol Al2O3
2 mol of Al 1 mol AL2O3
4.60 mol of Al X
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Sa.Ex. 3.18: 18.1 g of NH3 and 90.4g of CuO reacted NH3 + CuO N2 + Cu + H2O
1) Which is the limiting reactant?2) How many grams of N2 will be formed?
2NH3 + 3CuO N2 + 3Cu + 3H2O1) Balanced the equation
2) Find moles of the two reactants
Dr. Ali Bumajdad
3) Identify the limiting reactant (How?)
1.06
2= 0.53
1.14
3= 0.38
limiting
limiting18.1g ?
18.1 g
17.03 g/moln = = 1.06 mol
90.4g
90.4 g
79.55 g/moln = = 1.14 mol
2NH3 + 3CuO N2 + 3Cu + 3H2O
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Dr. Ali Bumajdad
3) Use the coefficient ratio between the limiting reactant
and the required substance to find out the number of mole
of the required substance
= 0.380 mol N2
4) Convert moles of required substance into masses
m N2 =n ×M.m.= 0.380×28.0 = 10.6 g N2M.m. n =
m (1)
3 mol of CuO 1 mol N2
1.14 mol of Al X
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Dr. Ali Bumajdad
•Theoretical yield, actual yield and percentage yield
•Mass of product•Maximum yield•Calculated using stiochiometry
•Mass of product•Known by experiment•Never more that theoretical yield
% Yield = Actual Yield
Theoretical Yieldx 100
(7)
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Sa.Ex. 3.19: 8.60 kg of H2 and 68.5kg of CO reacted H2 + CO CH3OH
1) Theoretical yield?2) % yield if the actual yield = 3.57 × 104 g CH3OH
1) Balanced the equation
2) Find moles of the two reactants
Dr. Ali Bumajdad
3) Identify the limiting reactant (How?)
4.27×103
2= 2135
2.44×103
1= 2440
limiting
limiting
2H2 + CO CH3OH
8.60 ×103g ?
8.60 ×103g
2.016 g/moln = = 4.27×103 mol
68.5 × 103g
68.5 × 103g28.02 g/mol
n = = 2.44×103 mol
2H2 + CO CH3OH
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3) Use the coefficient ratio between the limiting reactant
and the required substance to find out the number of mole
of the required substance
= 2135 mol CH3OH
4) Convert moles of required substance into masses
m CH3OH =n ×M.m.= 2135×32.04 = 6.86×104 g CH3OH
M.m. n =
m (1)
2 mol of H2 1 mol CH3OH4.27×103 mol of H2 X
% yield = 6.86×104 g
3.57 × 104 g= 52.0%× 100
Dr. Ali Bumajdad
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