2NaOH + FeCl2-----> 2NaCl + Fe(OH)2
a) Ta có
n\(_{NaOH}=\frac{4}{40}=0,1mol\)
Theo pthh
n\(_{FeCl2}=\frac{1}{2}n_{NaOH}=0,05\left(mol\right)\)
x=C\(_M\left(FeCl2\right)=\frac{0,05}{0,1}=0,5\left(M\right)\)
Theo pthh
n\(_{\downarrow}=\frac{1}{2}n_{NaOH}=0,05\left(mol\right)\)
m\(_{\downarrow}=0,05.90=4,5\left(g\right)\)
Chúc bạn học tốt