Giải thích các bước giải:
a.
PTHH:
$Fe+S\to FeS\ (1)$
$Fe+2HCl\to FeCl_2+H_2\ (2)$
$FeS+2HCl\to FeCl_2+H_2S\ (3)$
b.
$n_{Fe}=\dfrac{11,2}{56}=0,2\ (mol)$
$n_S=\dfrac{3,2}{32}=0,1\ (mol)$
Xét: $n_{Fe}=0,2\ (mol)>n_S=0,1\ (mol)$
$⇒$ Sau phản ứng Fe dư
$⇒n_{Fe\ pu}=n_{FeS}=n_S=0,1\ (mol)$
$⇒n_{Fe\ du}=0,2-0,1=0,1\ (mol)$
Theo PTHH (2): $n_{Fe\ du}=n_{H_2}=0,1\ (mol)$
Theo PTHH (3): $n_{FeS}=n_{H_2S}=0,1\ (mol)$
$⇒V=(0,1+0,1).22,4=4,48$ (lít)
c.
$H_2S+Pb(NO_3)_2\to PbS+2HNO_3$
BTNT S: $n_{PbS}=n_{H_2S}=0,1\ (mol)$
$⇒m_{PbS}=0,1.239=23,9\ (gam)$