What is the percent yield if only $26\mathrm{.}4gFeC{l}_3$ were obtained?

$31\mathrm{.}9\%$

$65\mathrm{.}8\%$

$50\mathrm{.}9\%$

$23\mathrm{.}8\%$

$S=32\frac{g}{m}ol,H=1\frac{g}{m}ol,Fe=56\frac{g}{m}ol,Cl=35\mathrm{.}5\frac{g}{m}ol$