1 gram molar mass of any gas at STP i.e, standard temperature 273K
and standard pressure 1 bar (760 mm of Hg) occupies 22.4 litres known as
gram molar volume.
∴ 2.0g of hydrogen occupies 22.4 litres at STP.
10.0g of hydrogen occupies ........? litres at STP.
10.0g x 22.4 litres/ 2.0 liters= 112 liters
2 g of hydrogen i.e, 1 mole of H2 contains 6.02x10 23
(NA) molecules
10 g of hydrogen
contain ....................?
10.0g x 6.02x10 23 molecules/
2.0g
= 30.10 x 10 23 molecules
= 3.01 x 10 24 molecules
Eg 3: Calculate the volume and No. of molcules of CO2
liberated at STP. If
50 g. of CaCO3
is treated with dilute hydrochloric acid which contains 7.3
g of dissolved HCl gas.
The Chemical equation for the above the reaction is
CaCO 3(S)
+ 2HCl (aq)
=> CaCl
2 (aq)
+ H 2
O (1)
+ CO2(g)
As per the metric staichio equation 100g of CaCO3
reacts with 73g of HCl
to liberate 44 g of CO2
In the above problem the amount of CaCO3
taken is 50g and HCl availble
is 7.3g.
100g of CaCO3
require 73g of HCl and 50g of CaCO3
required 36.5g of
HCl but, only 7.3g of HCl is available.
Hence the product CO2
formed depends only on the amount of HCl which
is in the least amount but not on the amount of CaCO3
which is an excess.
The reactant available in less amount is called limiting reagent as it limits
the amount of product formed.