Chemical Formula & Equation

1. STOICHIOME
TRY
1

2. Example 1:
2Mg + O2 2MgO
0.025 mol 0.025 mol
What is the mass of the oxide formed if 0.6 g of
magnesium is burnt in excess oxygen?
Number of moles of Mg = = 0.025 mol
0.6
24
2

3. Stoichiometry:
From the equation :
2 moles of magnesium, Mg produce
2 moles of magnesium oxide, MgO
From the calculation :
0.025 mole of magnesium, Mg produce
0.025 mole of magnesium oxide, MgO
Therefore, mass of MgO Mass = 0.025 x 40
= 1.0 g
n =_mass___
molar mass
3

4. Example 2:
CaCO3+ 2HCl CaCl2 + H2O+ CO2
0.05 mol 0.05 mol
What is the volume of carbon dioxide released when 5 g
of calciumcarbonate reacts with hydrochloric acid at
roomcondition?
Number of moles of CaCO3 = =
5.0
100
0.05 mol
4

5. Stoichiometry:
From the equation :
1 mole of calcium carbonate, CaCO3 produce
1 mole of carbon dioxide, CO2
From the calculation :
0.05 mole of calcium carbonate, CaCO3 produce
0.05 mole of carbon dioxide, CO2
Therefore, volume of CO2: Volume = 0.05 x 24
= 1.2 dm¯³
n =_volume__
molar volume
5

6. Example 3:
5.0 g of hydrated copper(II) sulphate, CuSO4.5H2O is
heated to remove its water vapour. Calculate the
mass of watervapour removed.
CuSO4.5H2O CuSO4 + 5H2O
Number of moles of CuSO4.5H2O = =
6

7. Stoichiometry:
From the equation :
..............................................................................................
..............................................................................................
...........................................................................................
From the calculation :
……………………………………………………………………
……………………………………………………………………
……………………………………………………………………
Therefore, mass of water vapour : Mass =
=
7

8. Example 4
3.1 g of sodiumoxide is formed when the sodium
metal is burnt completely in oxygen at STP.
Calculate the volume used.
4Na + O2 2Na2O
Number of moles of Na2O = =
8

9. Stoichiometry:
From the equation :
…………………………………………………………
…………………………………………………………
…………………………………………………………
From the calculation :
.………..
…………………………………………………………
…………………………………………………………
………………………………………………
Therefore, volume of oxygen:
9

10. Stoichiometry:
From the equation :
…………………………………………………………
…………………………………………………………
…………………………………………………………
From the calculation :
.………..
…………………………………………………………
…………………………………………………………
………………………………………………
Therefore, volume of oxygen:
9