Calculate the volume of `H_(2)` liberated at `27^(@)C` and 760 mm of Hg pressure by action by 0.6 g magnesium with excess of dil HCl.

Explanation

Chemical equation is
Mg `Mg -: 2HCl to MgCl_(2)+H_(2)`
24 gms of Mg `to 1` " mole of"`H_(2)` at STP
=22.4 lit at STP
0.6 gms. Of Mg `to` ?
`(0.6xx22.4)/(24)=0.56" lit"=560 " ml"`
Given conditions
`P_(1)=760` mm
`V_(1)=?`
`T_(1)=27^(@)C=300" k"`
STP conditions
`P_(2)=760` mm
`V_(2)=560` " ml"
`T_(2)=0^(@)C=273 " k" `
Formula
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`(760 xx V_(1))/(300)=(760 xx 560)/(273)`
`V_(1)=(560 xx 300)/(273)=615.4 " ml"`
`:. ` Volume of `H_(2)` liberated at `27^(@)C` and 760 mm of presence = 615.4 ml = 0.6154 Lit.