\(n_{Al_2O_3}=\dfrac{2,04}{102}=0,02\left(mol\right)\)
\(n_{HCl}=0,4V_1\left(mol\right)\)
\(n_{NaOH}=1,2V_2\left(mol\right)\)
Ta có: V1 + V2 = 0,5 (1)
TH1: HCl dư.
PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
Theo PT: nHCl = 6nAl2O3 + nNaOH
⇒ 0,4V1 = 0,02.6 + 1,2V2 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}V_1=0,45\left(l\right)\\V_2=0,05\left(l\right)\end{matrix}\right.\)
TH2: NaOH dư.
PT: \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
Theo PT: nNaOH = nHCl + 2nAl2O3
⇒ 1,2V2 = 0,4V1 + 0,02.2 (3)
Từ (1) và (3) \(\Rightarrow\left\{{}\begin{matrix}V_1=0,35\left(l\right)\\V_2=0,15\left(l\right)\end{matrix}\right.\)
