Question

Calculate the volume ,mass and number of molecules of H2 liberated when 230g of sodium reacts with excess of water at STP.
(at.mass of Na=23U,O=16U,H=1U

Answer 1

2Na+2H₂O⁻⁻⁻⁻2NaOH+H₂
so 46 g of Na reacts with water to give 22.4 L of H₂
so 230g of Na reacts with water to give 22.4X5=112 L of H₂
112 L of H₂=112/22.4=5 moles of H₂
mass of H₂ released=5X2=10g
number of molecules=5x6.023x$10^{23}$
                                     =30.115X$10^{23}$

Answer 2

Answer:

Explanation:

The balanced equation for the above reaction is as follows :

2Na (s) + 2H2O  + 2NaOH (aq) + H2 (g)

(2x23 )U + 2(2x1+1x16)U --> 2(23+16+1)U + (2x1)U

46 U + 36 U --> 80 U + 2 U

or 46 g + 36 g --> 80 g + 2 g

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  / 2.0g = 112 litres

                                   

2 g of hydrogen i.e, 1 mole of H2 contains 6.02x1023 (NA) molecules

10 g of hydrogen contain ....................?

10.0g x 6.02x1023 molecules  / 2.0g

= 30.10 x 1023 molecules

= 3.01 x 1024 molecules