The balanced equation is
`underset(24g)underset(1"mol")(Mg)+2HCl to MgCl_(2)+underset(22.4"litre of NTP")underset(1"mol")(H_(2))`
24g of Mg liberates hydrogen=22.4litre
1.2g Mg will libertes hydrogen=`(22.4)/(24)xx1.2=1.2"litre`
Volume of hydrogen under give condition can be calculated by applying `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`{:(,P_(1)=760mm,P_(2)=760mm),(,T_(1)=273K,T_(2)=(27+273)=300K),(,V_(1)=1.12"litre",V_(2)=?):}`
`V_(2)=(760xx1.12)/(273)xx(300)/(760)=1.2308"litre"`