Gọi a, b, c là mol mỗi KL trong 0,5 mol
\(\Rightarrow a+b+c=0,5\left(1\right)\)
Bảo toàn e: \(2b+3c=0,4.2+0,8\left(2\right)\)
Trong 47,6g X có ka, kb, kc mol mỗi KL
\(\Rightarrow k.\left(64a+56b+27c\right)=47,6\left(+\right)\)
\(n_{Cl2}=1,3\left(mol\right)\)
Bảo toàn e:\(k\left(2a+3b+3e\right)=1,3.2=2,6\left(++\right)\)
\(\left(+\right)+\left(++\right)\Rightarrow k=\frac{47,6}{64a+56b+27c}=\frac{2,6}{2a+3b+3c}\)
\(\Rightarrow2,6\left(64a+56b+27c\right)=47,6\left(2a+3b+3c\right)\)
\(\Rightarrow71,2a+2,8b-72,6c=0\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow a=c=0,2;b=0,1\)
\(\%Fe=\frac{0,1.56.100}{0,2.64+0,1.56+0,2.27}=23,53\%\)
