4)
дано
m(Fe2O3) = 24 g
W() = 3.4%
-------------------------
m(Fe)-?
m(Fe2O3) = 24 - (24*3.4% / 100% ) = 23.184 g
23.184 X
Fe2O3+2Al-->2Fe+Al2O3 M(Fe2O3) = 160 g/mol , M(Fe) = 56 g/mol
160 2*56
X = 23.184 * 2* 56 / 160 = 16.23 g
ответ 16.23 г
5)
FeCL3+3NaOH-->Fe(OH)3↓+3NaCL
бурый осадок