\(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)
0,2 0,1
\(n_{H_2}=\dfrac{2,128.100}{95}:22,4=0,1\left(mol\right)\)
\(m_{Na}=0,2.23=4,6\left(g\right)\)
\(m_{Fe}=10,2-4,6=5,6\left(g\right)\)
\(\%Na=\dfrac{4,6}{10,2}.100\%\approx45,1\%\)
\(\%Fe=\dfrac{5,6}{10,2}.100\%=54,9\%\)