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STOICHIOMETRY OF CHEMICAL REACTION
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● Interconvert masses, moles and formulas
● Determine formula from composition
● Perform calculations of purity of substances
● Interpret balanced chemical equations to calculate the moles of reactants and
products involved in each of the reactions
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● Compare the amount of substance actually formed in a reaction(actual yield) with the
predicted amount (theoretical yield) and determine the percentage yield
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The Word “Stoichiometry is derived from the Greek word stoicheion, which means measure
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THE MOLES CONCEPT
Moles is defined as the amount of substances that contains as many entities (atoms,
molecules, ion or other particles) as there are atoms in exactly 12
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e 1 mole=6
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,The mole concept, together with Avogadro number, provides important relationships among
the extensive properties! Mass, volume, number of moles, atoms, molecules, ions of any
given substance
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9g of iron metal contain?(Fe=55
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85g
X mole of Fe atoms=136
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9/55
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451mole Fe atoms
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The mass of one mole of Oxygen is 32
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e 1 mole of oxygen=32
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0g
X= 1×40÷32= 1
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1mol of Oxygen contains 6
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02x10^23 Oxygen molecules
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02×10^23÷32
X=7
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1 oxygen molecule contains 2 oxygen atoms
but 1 molecule of oxygen=32g
32g of oxygen= 6
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0g of Oxygen=X molecules
X= 6
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5×10^23
2 Oxygen atoms= 1 Oxygen molecules
Oxygen atoms= ½ Oxygen molecules
X= 1×7
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52×10^23
X=1
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How many atoms are contained in 2
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02×10^23 Fe
atoms
2
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451×6
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476×10^24 Fe atoms
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85g/mol)
solution
1 mole of Fe atoms=55
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85g Fe atoms= 6
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02×10^23)
Xg Fe atoms= 1 mol Fe atom
X=55
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02×10^23
X= 9
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0million SO2 molecules (Fwof SO2 =64
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02×10^23 SO2 molecules and 6
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1g(Since
1molSO2=64
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1÷6
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06×10^-15gSO2
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6g of Ammonium sulfate
(NH)2So4(FW of (NH4)2SO4 = 132
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1g
X mole of (NH4)2SO4 molecules= 39
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6/132
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30moles (NH4)2SO4
But 1 mo l(NH4)2SO4 molecules = 6
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0
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3×6
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81×10^23 molecules
But 1(NH4)2SO4 molecule = 8H atoms

1
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81×10^23×8
Z= 1
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COMPOSITION OF COMPOUNDS
Examples
● What mass of chromium, Cr is contained in 35
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(FW of
(NH4)2Cr2O7 = 252g/mol)
Solution
252g of (NH4)2Cr2O7 contained 104gCr
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8g of (NH4)2Cr2O7 will contain Xg , Cr
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77gCr
Alternatively,
I mol(NH4)2Cr2O7= 252g(NH4)2Cr2O7
Xmole(NH4)2Cr2O7=35
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8/252
X=0
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142mol(NH4)2Cr2O7=Y mol of Cr atoms
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142×2= 0
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0gCr
0
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284×52
Z=14
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●

●

What Mass of Potassium chlorate KClO3 would contain 40
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6g/mol)
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6g KClO3 contained 48g of O
XKClO3 will contain 40g of O
X=122
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16g of KClO3
Alternatively
I mol of O atoms =16
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0g of O atoms
X=1×40/16=2
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5mol of O atoms
Y=1×2
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833mol of KClO3
Also 1 mol of KClO3= 122
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833mol of KClO3=Zg KClO3
Z=0
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6
Z=102
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What mass of Sulphur dioxide,SO2, would contain the same mass of Oxygen as is
contained in 33
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1g/mol, FW of
AS2O5=229
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8g AS2O5 contain 80g of O

33
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7×80÷229
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73gO
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1g SO2 contains 32g of O
YgSo2 will contains 11
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1 × 11
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50g of SO2
● 67
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xH2O was heated to drive away all its
Water of crystallization
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4g what is the formula of
hydrate
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2g/mol)
Solution
Imol CaSo4 = 136
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4g CaSo4
Z= 1×53
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2=0
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Also 1 mol H20 =18gH2O
Mol H2O=14
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5-53
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1g)
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1×1/18= 0
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783×1/0
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REACTION STOICHIOMETRY
Mole, mass and volume relationships
1
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20mol of CH4
● the volume of steam produced at STP
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20molCH4 =Xg of O2
X=1
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8g of O2
● Since 1 mol of O2 =32g of O2
X mole of O2 =76
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8/32
At STP, 1 mol of Oxygen= 22
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4mol of O2 =Ydm^3
Y=2
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4/1
Y= 53
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what mass of P4 reacts with 1
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M of P =31)
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5mol of O2
X= 1×1
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3mol of P4
1 mol of P4 = 124
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3mol of P4 = YgP4
Y= 0
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20gP4
CONCENTRATION
● Calculate the mass of nickel (11) sulfate, 6
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Xg= 6
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0gNiSO4
● Calculate the mass of NiSO4 present in 200ml of a 6
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06g/mol at 25 degree centigrade
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06g NiSO4 is present in 1ml
XgNiSO4 is present in 200 ml
X=1
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0g of NiSO4 is present in 100g of solution
Xg of NiSO4 is present in 212g of Solution
X=6×212/100
X=12
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● calculate the molarity of a solution that contains 3
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0litre of solution (FW
of HCl=36
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5gHCl
X mole Hcl = 3
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65/36
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1mol HCl
0
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0liter
Ymol HCL will be present in 1
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1×1/2
Y=0
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5liter of a 0
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3g)
Solution
From the molar concentration
0
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0Liter
Xmole Ba(OH)2 is present in 2
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06 × 2
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15mol Ba(OH)2
But 1 mol of Ba(OH)2 = 171
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15mol Ba(OH)2 =Yg
Y= 0
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3
Y= 25
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324M H2SO4 Solution
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324mol H2SO4 is present in 1000 ml solution
(Since 1L= 1000ml)
X mol H2SO4 present in 500ml
X= 0
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162mol of H2SO4
(b) 1 mol
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162mol of H2SO4 = Xg we
X=0
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9g H2SO4
VOLUMETRIC ANALYSIS
Examples
● Calculate the volume in Liters of a 0
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792g of Na2Co3 according to the equation
H2SO4 + Na2CO3 — Na2SO4 + CO2 + H2O
(H=1, S=32,O=16,Na =23,C=12)
Solution
1 mol Na2CO3 = 106gNa2CO3
X mol Na2CO3 =2
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792/106
X= 0
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0
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026mol Na2CO3
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324mol H2SO4 is present in 1 liter
0
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026×1/0
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08 Liter
● Find the volume in Liters of a 0
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0ml
of 0
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0
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X mol H2SO4 will be present in 40
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505×40/1000
X=0
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From the balanced equation above
1 mol H2SO4 react with 2 mol NaOH
0
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Y= 0
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04mol NaOH
0
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04mol NaOH will be present in X liters
Z = 0
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505
Z= 0
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8g of an acid Hx per dm^3
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10mol/dm^3 of Sodium hydroxide
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00Cm^3 of A neutralized 25
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Calculate the
(a) Molar concentration of Hx
(b) Molar mass of Hx
(c) Atomic mass of X
● Solution
(a) NaOH + Hx —- NaX + H2O
0
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X
Mol NaOH will be present in 25 cm^3
X= 24×0
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0025mol NaOH
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0025mol NaOH will react with 0
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If 0
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00cm^3 (titre value)
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0025/24
Y=0
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(b) If 3
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104mol Hx is in 1 dm^3, then
0
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8g Hx
1mol Hx = ZgHx
Z = 1×3
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104
Z= 36
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5-1
= 35
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S(2017) chemistry: The molecular Nature of Matter and change (8th ed) McGrawHill Education

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