\(n_{NO}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
\(\begin{matrix}\overset{0}{Mg}\rightarrow\overset{+2}{Mg}+2e\\\overset{0}{Fe}\rightarrow\overset{+3}{Fe}+3e\\\overset{+5}{N}+3e\rightarrow\overset{+2}{N}\end{matrix}\)
Bảo toàn e:
\(2n_{Mg}+3n_{Fe}=3n_{NO}=1,05\left(1\right)\)
Lại có \(24n_{Mg}+56n_{Fe}=15,6\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}n_{mg}=0,3\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,3.24=7,2\left(g\right)\Rightarrow\%m_{Mg}=\dfrac{7,2}{15,6}.100\%=46,15\%\)
\(\Rightarrow m_{Mg}=0,15.56=8,4\left(g\right)\)
\(\%m_{Fe}=100\%-46,15\%=53,85\%\)
Bảo toàn N:
\(n_{HNO_3}=n_N=2n_{Mg\left(NO_3\right)_2}+3n_{Fe\left(NO_3\right)_3}+n_{NO}\)
\(=2n_{Mg}+3n_{Fe}+n_{NO}\)
\(=2.0,3+3.0,15+0,35=1,4\left(mol\right)\)
\(C_M=\dfrac{1,4}{0,2}=7M\)
