Answer
See answer below.
Work Step by Step
a) $2\ Fe(s)+3\ Cl_2(g)\rightarrow 2\ FeCl_3(aq)$
b) Number of moles of iron:
$10.0\ g\div 55.845\ g/mol=0.179\ mol$
Mass of chlorine:
$0.179\ mol\times3/2\times 70.90\ g/mol=19.04\ g$
Mass of chloride:
$0.179\ mol\times 162.2\ g/mol=29.03\ g$
c) $18.5/29.03\times 100\%=63.72\%$
d) Number of moles of chlorine:
$10.0\ g\div 70.90\ g/mol=0.141\ mol$, it's the limiting reactant
Mass of chloride:
$0.141\ mol\times 2/3\times 162.2\ g/mol=15.25\ g$
