The decomposition of hi has a rate constant of 1.80 × 10-3 m-1s-1. how much hi remains after 8.50 min if the initial concentration of hi is 4.78 m?

By  use  second   order  integrated  integrated  law 
1/(A)t  =Kt  + 1/(A)o
k= constant
t=  time  in  second=  8.50  x60=  510  sec
(  A)o=  initial  concentration
(A)t  =final   concentration
1/(A)t  =(1.80  x10^-3) (510)  +1  /4.78
1(A)t=  1.127  multiply  both  side  by  (A)t
1=1.127 (A)t    divide  both  side by  1.127
(A)t=  0.887  M

Reaction type : double replacement hcl + lioh = h2o + licl