\(n_{NaOH}=2.0,3=0,6\left(mol\right)\)
\(PTHH:FeCl_x+xNaOH\rightarrow Fe\left(OH\right)_x+xNaCl\)
pt:__\(56+35,5x\left(g\right)\)__\(x\left(mol\right)\)_________________
pứ:____32,5(g)_____0,6(mol)__________________
Áp dụng định luật tỉ lệ:
\(\frac{56+35,5x}{32,5}=\frac{x}{0,6}\Rightarrow x=3\)
\(\rightarrow CTHH:FeCl_3\)