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 = 16U, and H = 1U)
the balanced equation for the above reaction is
`{:(2Na_((s))+2H_(2)O_((l)),,to2NaOH_((aq)) +H_(2(g))+H_(2(g))uparrow),((2xx23)U+ 2(2xx1 +1 xx 16)U,,to2(23+16+1)U+(2xx1)U),(46 U + 36U,,to80U + 2U),(or46g + 36g ,, to 80 g + 2 g):}`

Explanation

As per the balanced equation :
46 g of Na gives 2g of hydrogen
230 g of Na gives ________ ? G of hydrogen
`(230 g xx 2g)/(46 g) = 10g` of hydrogen
1 gram molar mass of any gas at STP i.e, standard temperature 273 K and standard pressure 1 bar, occupies 22.4 litres known as gram molar volume.
`therefore` 2.0 g of hydrogen occupies 22.4 litres of STP.
10. 0 g of hydrogen occupies ............. ? litres of STP.

`(10.0 g xx 22.4 "litres")/(2.0 g) = 112` litres
2 g of hydrogen i.e, 1 moles of `H_(2)` contains `6.02 xx 10^(23) (N_(0))` molecules
10 g of hydrogen contain ................ ?
`(10.0 g xx 6.02 xx 10^(23) "molecules") /(2.0 g)`
` = 30.10 xx 10^(23)` molecules
` = 3.01 xx 10^(24)` molecules