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Answer 1

\(a,PTHH:2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ b,n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ \Rightarrow n_{FeCl_3}=n_{Fe}=0,1(mol)\\ \Rightarrow m_{FeCl_3}=0,1.162,5=16,25(g)\\ c,C_{M_{FeCl_3}}=\dfrac{0,1}{0,1}=1M\)