`A_(2)+2B_(2)rarrA_(2)B_(4)`
`(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4)`
Two substance `A_(2) & B_(2)` react in the above manner when `A_(2)` is limited it gives `A_(2)B_(4)` in excess gives `A_(3)B_(4^(.) A_(2)B_(4)` can be converted to `A_(3)B_(4)` when reacted with `A_(2)`. Using this information calculate the composition of the final mixture when the mentioned amount of `A % B` are taken `:c`
`(a) 4` mole `A_(2) & 4`mole `B_(2)`
`(b) (1)/(2)` moles `A_(2) & 2` moles `B_(2)`
`(c ) 1.25` moles `A_(2) & 2` moles `B_(2)`

Answer

`(a) A_(3)B_(4)=2&A_(2)=1 ; (b) A_(2)B_(4)=(1)/(4)&B_(2)=1(c )A_(2)B_(4)=0.5&A_(3)B_(4)=0.5`

Explanation

`{:((a)A_(2)+2B_(2)rarrA_(2)B_(4)),("Initial" 4" "4),("After"4-2" " "4-4 "2),(" "2" "0" "2):|(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4)`
`2A_(2)B_(4) " "+" "A_(2)" "rarr " "2A_(3)B_(4)`
`2" "2`
`2-2" "2-1" "2`
`1" "2`
`therefore A_(2)=1,A_(2)B_(4) =2`
`{:(A_(2)+2B_(2)rarrA_(2)B_(2)),("initial"(1)/(2)" "2),("After"1.25-1-" "1),(0.25 " "-" "1):}:|{:(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4):}`
`therefore A_(2)B_(4)=0.5,B_(2)=1`
`{:(A_(2)+2B_(2)rarrA_(2)B_(4)),("Initial"1.25" "2),(0.25" "-" "1):}:|(3)/(2)A_(2)+2B_(2)rarrA_(3)B_(4)`
`2A_(2)B_(4)+A_(2)rarr2A_(3)B_(4)`
`1" "0.25`
`1-0.5" "-" "0.5`
`therefore A_(2)B_(4)=A_(3)B_(4)=0.5`