Question

Answer 1

Bài 2:

a) 

\(n_{MgCl_2}=0,15.1=0,15\left(mol\right)\)

PTHH: MgCl₂ + 2NaOH --> Mg(OH)₂\(\downarrow\) + 2NaCl

              0,15---->0,3--------->0,15

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,15.58=8,7\left(g\right)\)

b) \(C_{M\left(dd.NaOH\right)}=\dfrac{0,3}{0,2}=1,5M\)

c)

PTHH: Mg(OH)₂ --t^o--> MgO + H₂O

              0,15---------->0,15

=> m = 0,15.40 = 6 (g)