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

Explanation

The balanced equation for the above reaction is
`{:(2Na_((s)),+,H_(2)O,rarr,2NaOH_((aq)),+,H_(2(g))),((2xx23)U,+,(2xx1+1xx16)U,rarr,(2xx23+1xx1+1xx16)U,+,(2xx1)U),(46U,+,36U,rarr,80U,+,2U),(46gm,+,36gm,rarr,80gm,+,2gm):}`
A) Mass
46 g of Na gives 2g of hydrogen
230 g of Na gives `overset(?)("_____________")`g of hydrogen.
`x=((230)gmxx2gm)/(46gm)=10g` of hydrogen.
B) Volume
2.0 g of hydrogen occupies 22.4 litres at STP.
10.0 g of hydrogen occupies `overset(?)".........."` litres at STP.
`x=(10gmxx"22.4 litres")/(2gm)="112 litres"`
C) Number of molecules
2g of hydrogen i.e., mole of `H_(2)` contains `6.02xx10^(23)` number of molecules.
10g of hydrogen i.e., 5 mole of `H_(2)` contains ............?............. number of molecules.
`x=(10g xx6.02xx10^(23))/(2gm)=30.10xx10^(23)` molecules