Answer
See answer below.
Work Step by Step
a) $FeCl_2(aq)+Na_2S(aq)\rightarrow FeS(s)+2\ NaCl(aq)$
b) Number of moles of $FeCl_2$:
$40.\ g\div 126.74\ g/mol= 0.316\ mol$
Number of moles of $Na_2S$:
$40.\ g\div 78.04\ g/mol= 0.513\ mol$
$FeCl_2$ is the limiting reactant.
c) $0.316\ mol\times 87.91\ g/mol= 27.78\ g$
d) $(0.513-0.316)\ mol\times 78.04\ g/mol= 15.37\ g$
e) $0.513\ mol\times 126.74\ g/mol=65.02\ g$
