Answer
See the answer below.
Work Step by Step
a) Reacted concentration: x
$K_c=[NH_3][H_2S]$
$1.8\times10^{-4}=x^2$
$x=0.0134\ M$
$[NH_3]=[H_2S]=0.0134\ M$
b) Reacted concentration: x
$K_c=[NH_3][H_2S]$
$1.8\times10^{-4}=(0.02+x)x$
$x=0.0067\ M$
$[NH_3]=0.0267\ M, [H_2S]=0.0067\ M$
