Cho 100 ml dung dịch gồm MgCl2 1M và AlCl3 2M

Đáp án C

$n_{MgC{l}_2}=0,1mol$; $n_{AlC{l}_3}=0,2mol$  ; $n_{Ba{\left(OH\right)}_2}=0,425mol$

Do đó: $n_{M{g}^{2+}}=0,1mol$ ;$n_{A{l}^{3+}}=0,2mol$ ;$n_{O{H}^{-}}=0,85mol$

Phương trình hóa học:         $M{g}^{2+}+2O{H}^{-}\to Mg{\left(OH\right)}_2\downarrow$

                                          0,1 → 0,2    →         0,1                       mol

                                     $A{l}^{3+}+3O{H}^{-}\to Al{\left(OH\right)}_3\downarrow$

                                        0,2 → 0,6    → 0,2                       mol

                                     $Al{\left(OH\right)}_3\downarrow +O{H}^{-}\to Al{O}_2^{-}+2H_{2}O$

                                      0,05 ←0,05                                      mol

$\to n_{Al{\left(OH\right)}_3}=0,2-0,05=0,15mol$; $m_{keáttuûa}=m_{Mg{\left(OH\right)}_2}+m_{Al{\left(OH\right)}_3}=0,1.58+0,15.78=17,5gam$ .