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Theo đề bài, ta có: \(\left\{{}\begin{matrix}m_{Fe\left(NO3\right)3}=6,05.\left(56+\left(14+3.16\right).3\right)=1464,1\left(g\right)\\\%Fe=\dfrac{56}{\left(56+\left(14+3.16\right).3\right)}.100\%\approx23,14\%\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=1464,1.23,14\%=338,79274\left(g\right)\)
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