nAl(OH)3= 15.6/78=0.2 mol
PI : nH2SO4= 0.2 mol
2Al(OH)3 + 3H2SO4 --> Al2(SO4)3 + 6H2O
Bđ: 0.2_________0.2
Pư : 1/15________0.2_________1/15
Kt: 2/15_________0___________1/15
mAl2(SO4)3= 1/15*342= 22.8g
PII:
nNaOH = 0.05 mol
NaOH + Al(OH)3 --> NaAlO2 + 2H2O
Bđ: 0.05_____0.2
Pư: 0.05_____0.05______0.05
Kt: 0________0.15______0.05
mNaAlO2= 0.05*82=4.1g
a) \(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)
\(m_{Al\left(OH\right)_3}=\frac{1}{2}\times15,6=7,8\left(g\right)\)
\(\Rightarrow n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1\left(mol\right)\)
PTHH: 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Theo PT: \(n_{Al\left(OH\right)_3}=\frac{2}{3}n_{H_2SO_4}\)
Theo bài: \(n_{Al\left(OH\right)_3}=\frac{1}{2}n_{H_2SO_4}\)
Vì \(\frac{1}{2}< \frac{2}{3}\) ⇒ H2SO4 dư
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=\frac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\times342=17,1\left(g\right)\)
b) \(n_{NaOH}=0,05\times1=0,05\left(mol\right)\)
\(m_{Al\left(OH\right)_3}=\frac{1}{2}\times15,6=7,8\left(g\right)\)
\(\Rightarrow n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1\left(mol\right)\)
PTHH: Al(OH)3 + NaOH → NaAlO2 + 2H2O
Theo PT: \(n_{Al\left(OH\right)_3}=n_{NaOH}\)
Theo bài : \(n_{Al\left(OH\right)_3}=2n_{NaOH}\)
Vì \(2>1\) → Al(OH)3 dư
Theo Pt: \(n_{NaAlO_2}=n_{NaOH}=0,05\left(mol\right)\)
\(\Rightarrow m_{NaAlO_2}=0,05\times82=4,1\left(g\right)\)
