What volume of `H_(2)(g)` is produced by decomposition of `2.4 L NH_(3) (g)` ? .

Explanation

` NH_(3)(g)rarr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`
`{:(t=0,2.4L,0,0,),(t=oo0,,v_(1),v_(2),):}`
Gay-Lussac's law `2.4 : v_(1) :v_(2) = 1 : (1)/(2) : (3)/(2) rArr (2.4)/(v_(2)) = (2)/(3) rArr v_(2) = 3.6L` .