Balance equation
1.
2.
Group 1 metal + water  alkaline + H2
Metal + O2  metallic oxide

Oxidation number!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
-
4Al + 3O2  2Al2O3
Mg + Br2  MgBr2
2Fe + 3Cl2  2FeCl3
4Li + O2  2Li2O
Zn + F2  ZnF2
3Mg + N2  Mg3N2
3.
Displacement reaction
Metal
Reactivity series
Non-metal
F > Cl > Br > I
No one reactive element + less reactive compound
 less reactive element + more reactive compound
Oxidation number !!!!!!!!
-
Zn + AgNO3  ZnNO3 + Ag
Mg + AlCl3 
Cl2 + BaBr2 
A
B
C
D
Density
0.50
0.70
2.40
3.50
Conductivity ofelectricity
Yes
No
Yes
no
Hardness
Soft
Hard
Hard
Soft
Group1 metal is ‘A’
How many grams of 3.2mol of N2 gas?
14+14 * 3.2 = 89.6
What is the volume of 3.2 mol of N2 gas under r.p.t
3.2 * 24dm-3 = 76.8
How many mole of NaCl needed to prepare 1.3mole*dm-3 500cm3 solution
0.5 *1.3 = 0.65
What is the concentration when dissolve 0.6 mol of NaOH into water to make a 60cm3 solution?
0.6 / 0.06 = 10mol*dm-3
2g of NaHCO3 reacts with excess HCl (aq) calculate the volume of Co2(g)
NaHCO3 (s) + HCl(aq)  NaCl (aq) + H2O (l) + CO2(g)
1. 2 g/ 23+1+12+16*3 + 0.02 mol NaHCO3
2.
3.
Mole of CO2 = mole of NaHCO3
V Co2 = 0.02 * 24dm-3 = 0.48 dm-3
MOLE CALCULATION
1.
18 grams of limestone reacted with excess dilute hydrochloric acid. 3840cm^3 of carbon dioxide was
formed with r.t.p. work through the calculation to find the percentage purity of this sample of limestone,
which is impure calcium carbonate, CaCO3. Give yours to two significant figures. Ar value : C=12 , Ca=40 ,
O=16
CaCO3 + 2HCl  CaCl2 + CO2 + H2O
1.
2.
3.
4.
5.
6.
The molar mass of calcium carbonate: 100g/mol
Volume of CO2 in dm-3 = 3.84
Moles of CO2 = 3.84/24 = 0.16mol
Moles of CaCO3 in limestone sample = 0.16 mol
Mass of CaCO3 in limestone sample = 16g
Percentage purity = 16/18 * 100 = 88%
2. Methyl benzoate can be prepared by reacting methanol with benzoic acid
CH3OH (methanol) + C6H5CO2H (benzoic acid)  C6H5CO2CH3 + H2O
When 24.4g of benzoic acid is reacted with excess methanol, 25.84g of methyl benzoate is produced. Work
through the calculation to find the percentage yield of methyl benzoate. Ar value : C=12 , H=1 , O=16
1.
2.
3.
4.
5.
6.
Molar mass of benzoic acid = 122g/mol
Moles of benzoic acid = 24.4/122 = 0.2 mol
Moles of methyl benzoate expected (if 100% yield) = 0.2mol
Molar mass of methyl benzoate = 136g/mol
Mass of methyl benzoate expected (if 100% yield) = 0.2 * 136 = 27.2g
Percentage yield = 25.84/27.2 *100 = 95
3. A student reacts to 5.4g of aluminum with excess oxygen.
4Al + 3O2  2Al2O3
The mass of aluminum oxide produced is 8.67 g. Calculate the percentage yield.
Ar values: Al=27 , O=16
1.
2.
3.
4.
Mole of aluminum = 5.4/27 = 0.2mol
Mole of 2Al2O3 = 0.1mol
Mass of 2Al2O3 = 0.1*102 = 10.2
Yield = 8.67/10.2*100 = 85%