All elements can be divided on metals and not metals
In periodic table metals are on the left. Not metals are on the right.
Group number shows how many electrons are in the most outer orbit. These electrons are called valence electrons. For example, Na (sodium) is in the first group. It has one electron on the most outer orbit Na can easily give this electron to Cl (chlorine). Cl is in the seventh group. It has 7 electrons and it takes one electron from Na. As a result Na becomes a positive ions Na+ and Cl becomes negative ions Cl -. Ions with opposite charge form ionic bond.
Na and Cl atoms form ionic bonds. Ionic bonds usually form crystal structures. That is why salt is made of crystals.
Carbon C located in 4th group. It has four valence electrons. As a result C forms four covalent bonds with four atoms of Cl. C does not give its electrons to Cl. Carbon and chlorine share electrons. When atoms share electrons they form covalent bonds.
Oxides are produced when metals or not metals react with oxygen.
Oxygen located in 6th group and has 6 valence electrons. It tends to gain 2 more electrons to get complete octet and its valence is 2.
2Ca + O2 = 2CaO
In nature metal oxides exist in clay. Clay is a mixture of the oxides
1. NaNO3 = NaNO2 + O2
Let us write half reaction of the oxidation and reduction. Initially nitrogen has charge +5 And at the end of reaction it has charge +3. How do we calculate that? In NaNO3 oxygen has charge -2. Sodium has charge +1. The molecule of NaNO3 is neutral. It means that negative charges inside NaNO3 molecule must be equal to positive charges.
Na (+1) + O3 (-2 x 3) = 1 - 6 = - 5. Then nitrogen has to be + 5 to make the molecule neutral.
In NaNO2, nitrogen has charge +3. Nitrogen must receive 2 negative electrons to change its charge from +5 to +3. 5 + ( - 2) = 3 So we can write
N5+ + 2e = N3+ | 2
Oxygen initially has charge -2. At the end of reaction it becomes neutral and has charge 0.
So, we can write 2O2- - 4e = O2 | 4
Combine two half reactions and get:
N5+ + 2e = N3+ | 2
2O2- - 4e = O2 | 4
Since 2 and 4 can be divided by 2 and reduced to 1 and 2
N5+ + 2e = N3+ | 2 1
2O2- - 4e = O2 | 4 2
Now switch positions of 1 and 2
N5+ + 2e = N3+ | 2 1 2
2O2- - 4e = O2 | 4 2 1
From the above N5+ and N3+ should have coefficient 2 and oxygen should have coefficient 1.
2NaNO3 = 2NaNO2 + O2
Check the equation balance: from both sides of the equation we have 2 Na, 2 N, and 6 O. We are done!
Balance the following Redox equations:
2. Fe + H2SO4 = FeSO4 + H2
3. NO2 + H2O = HNO3 + HNO2
4. FeS2 + HNO3 = Fe(NO3)3 + H2SO4 + NO2
5. HgO = Hg + O2
6. AgNO3 + H2O = Ag + HNO3 + O2
7. Fe2O3 + H2 = Fe + H2O
8. H2O = H2 + O2
9. Fe2O3 + CO = Fe + CO2
10. KClO3 = KCl + O2
11. H2O2 = H2O + O2
12. HBr + H2O2 = Br2 + H2O
13. MnCO3 + KClO3 = MnO2 + KCl + CO2
14. H2S + SO2 = S + H2O
15. Sb + HNO3 = HSbO3 + NO2 + H2O
16. Al + CuCl2 = AlCl3 + Cu
17. Zn + CuSO4 = ZnSO4 + Cu
18. MnS + HClO3 = MnSO4 + HCl
19. H2S + FeCl3 = S + FeCl2 + HCl
21. Bi + HNO3 = Bi(NO3)3 + NO2 + H2O
22. PbS + HNO3 = PbSO4 + NO2 + H2O
23. C + HNO3 = CO2 + NO2 + H2O
24. FeSO4 + Br2 + H2SO4= Fe2(SO4)3 + HBr
25. Al + HCl = AlCl3 + H2
26. KMnO4 + SO2 + H2O = MnSO4 + H2SO4 + K2SO4
27. MnO2 + HCl = MnCl2 + H2O + Cl2
28. Cl2 + KOH = KCl + KClO3 + H2O
29. KMnO4 + NH3 = MnO2 + KOH + N2 + H2O
30. Mg + HNO3 = Mg(NO3)2 + NH4NO3 + H2O
If you have difficulties to balance these Redox reactions find answers and explanation in my tutorial
In reaction with water metals or metal oxides produce base:
2Na + 2H2O = 2NaOH + H2
CaO + H2O = Ca(OH)2
Bases dissociate in water and produce OH - ion.
Non metal oxides are NO2, SO3, P2O5
In reaction with water non metal oxides produce acids:
H2O + SO3 = H2SO4 - sulfuric acid
H2O + NO2 = HNO3 - nitric acid
H2O + CO2 = H2CO3 - Carbonic acid
Acids dissociate in water and produce proton of hydrogen H+
When an acid reacts with a base, a salt is produced
NaOH + HNO3 = NaNO3 + H2O
Calculate the percentage composition of NaNO3?.
First find molecular mass of NaNO3
23 (Na) + 12 (N) + 16*3 (O3) = 83
Molecular weigt of NaNO3 = 83
83 - 100%
23 (Na) - X% X = 23 * 100 /83 = 27.7% of Na
83 - 100%
12 (N) - X % X = 12 * 100/83 = 14.5% of N
83 - 100%
48 (O) - X % X = 48 * 100/83 = 57.8 % of O
Some salts are more soluble in water; some are less soluble or not soluble. When a not soluble salt is produced as a result of acid and base reaction a precipitate is formed.
Ca(OH)2 + H2CO3 = CaCO3 + 2H2O
CaCO3 is not soluble in water. A white precipitate is formed.
Salts may react with each other and new salts are produced:
All chemical reactions occur in equivalent proportions
1. Na2CO3 + Ca Cl2 = CaCO3 + 2NaCL
All compounds react with each other in certain proportions
In given reaction one mole of Na2CO3produces one mole of CaCO3.
Mole is MW(Molecular mass) in grams.
For Na2CO3 MW is 23 *2 + 12 + 48 = 106 g. = 1 mole
For CaCO3 MW is 40 + 12 + 48 = 100g. = 1 mole.
106 g produces 100g
10 g produces X g X= 10 * 100 / 106 = 9.4 g
2. Na2CO3 + Ca Cl2 = CaCO3 + 2 NaCL
100 ml g?
1 liter of I M solution of Na2CO3 contains 106 g
How many grams of Na2CO3 in 1 liter of 0.5 M solution?
1 M - 106 g
0.5 M - X g X = 0.5 M * 106 g / 1 M = 53 g
I liter of 0.5 M solution contains 53 grams
How many grams are in 100 ml?
1 liter - 53 g
0.1 liter - X g X = 0.1 * 53 /1 = 5.3 g
106 g of Na2CO3 produces 100 g of CaCO3
5.3 g of Na2CO3 produces X g of CaCO3 X= 5.3 * 100 / 106 = 5 g
My chemistry tutorial is focused on key chemistry topics: Oxides, Bases, Acids, Salts, Equivalent proportions, Acid Base reactions, Weigh and Volume problems, Equilibrium, Le Chatelier's Principle, Liquid Properties: Freezing and Boiling points, Balance Redox Reactions (30 examples with explanations), Stoichiometry (30 problems with answers and solutions). Study my tutorial and you will understand chemistry.