Question

calculate the volume ,mass and no. of molec. oh hydrogen libertated when 230g od Na rects with excess of water at STP
atomic mass of Na =23u.o=16u,h=1u

Answer 1

2Na + 2H2O → 2NaOH + H2 (↑)

we see 2 mole of Na form 1 mole of H2 gas

first of all we find out no of mole of Na
no of mole of Na = given wt/ molar wt
= 230/23 = 10
hence no of mole of Na = 10

so, no of mole of H2 liberated = 5 mole
volume of H2 liberated = mole x 22.4 L
= 5 × 22.4 = 112 L

mass of H2 liberated = mole x molar wt
= 5 × 2 = 10 g

Answer 2

2Na + 2H2O -----------→ 2NaOH + H2


46g of sodium form 2g of H2

now 23 g of sodium form 1g of H2

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230g of Na reacts with h2o form ................   10g of H2

1 mole of the H2 = 2 g

2g of the H2 contain 6.022x 10^-24  molecules

10g of the H2 contain = 6.022x 10^-24  x 10 g / 2g
                                   =    30.10 x 10^-24  molecules
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1 mole of H2 = 22.4 l
Now as 5 mole here of H2 then
its contain
=  5 x 22.4 l
=  112 l  
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