\(\begin{array} {l} a)\\ n_{NaOH}=\dfrac{8}{40}=0,2(mol)\\ 2NaOH+FeCl_2\to Fe(OH)_2\downarrow+2NaCl\\ n_{FeCl_2}=\dfrac{1}{2}n_{NaOH}=0,1(mol)\\ m_{dd\,FeCl_2}=\dfrac{0,1.127}{25,4\%}=50(g)\\ b)\\ n_{Fe(OH)_2}=n_{FeCl_2}=0,1(mol)\\ 4Fe(OH)_2+O_2\xrightarrow{t^o}2Fe_2O_3+4H_2O\\ n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe(OH)_2}=0,05(mol)\\ m=m_{Fe_2O_3}=0,05.160=8(g) \end{array}\)
\(a,n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: \(2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
0,2------->0,1-------->0,1
\(\rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
\(b,\) PTHH: \(4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^o}2Fe_2O_3+2H_2O\)
0,1------------------>0,05
\(\rightarrow m=0,05.160=8\left(g\right)\)
