\(n_{Al\left(OH\right)_3}=\frac{3,9}{78}=0,05\left(mol\right)\)
\(n_{Al_2\left(SO_4\right)_3}=0,2.0,2=0,04\left(mol\right)\)
Ta có
\(0,04.2>0,05\rightarrow\) Al(OH)3 tan 1 phần tạo NaAlO2
\(n_{NaAlO_2}=0,04.2-0,05=0,03\left(mol\right)\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)
____0,04_____0,24______________________________
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
__ 0,03_____0,03_________0,03_____________
\(m_{dd_{NaOH}}=\frac{0,27.40}{10\%}=108\left(g\right)\)
\(n_{Al_2\left(SO_4\right)_3}=0,04\left(mol\right) \Rightarrow n_{Al^{3+}}=0,08\left(mol\right)\\ n_{Al\left(OH\right)_3}=0,05\left(mol\right)\)
\(TH_1:Al\left(OH\right)_3\text{ không bị hòa tan }\\ Al^{3+}+3OH^-\rightarrow Al\left(OH\right)_3\\ \Rightarrow n_{OH^-}=3n_{Al\left(OH\right)_3}=0,15\left(mol\right)=n_{NaOH}\\ \Rightarrow m_{NaOH}=6\left(g\right)\Rightarrow m_{dd\text{ }NaOH}=60\left(g\right)\)
\(TH_2:Al\left(OH\right)_3\text{ bị hòa tan 1 phần}\)
Al3+ + 3OH- --> Al(OH)3
0,08__0,24_____0,08
Al(OH)3 + OH- ---> AlO2- + H2O
0,03______0,03
\(\Rightarrow m_{dd\text{ }NaOH}=108\left(g\right)\)
Đáp án:
108 g
Giải thích các bước giải:
nAl(OH)3=3,9/78=0,05 mol
nAl2(SO4)3=0,2x0,2=0,04 mol
ta có
0,04x2>0,05=> Al(OH)3 tan 1 phần tạo NaAlO2
nNaAlO2=0,04x2-0,05=0,03 mol
Al2(SO4)3+6NaOH->2Al(OH)3+3Na2SO4
0,04 0,24
Al(OH)3+NaOH->NaAlO2+2H2O
0,03 0,03 0,03
mdd NaOH=0,27x40/10%=108 g