1)
Gọi \(\left\{{}\begin{matrix}n_{BaO}=2a\left(mol\right)\\n_{Al_2O_3}=a\left(mol\right)\end{matrix}\right.\)
=> 306a + 102a = 48,96
=> a = 0,12 (mol)
\(\left\{{}\begin{matrix}\%m_{BaO}=\dfrac{0,24.153}{48,96}.100\%=75\%\\\%m_{Al_2O_3}=\dfrac{102.0,12}{48,96}.100\%=25\%\end{matrix}\right.\)
PTHH: \(BaO+H_2SO_4\rightarrow BaSO_4\downarrow\)
0,24--->0,24------>0,24
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,12----->0,36-------->0,12
\(n_{H_2SO_4\left(pư\right)}=0,24+0,36=0,6\left(mol\right)\Rightarrow n_{H_2SO_4\left(tt\right)}=0,6.140\%=0,84\left(mol\right)\)
=> \(m_{dd.H_2SO_4}=\dfrac{0,84.98}{43,12\%}=\dfrac{2100}{11}\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,12.342}{48,96+\dfrac{2100}{11}-0,24.233}.100\%=22,31\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,84-0,6\right).98}{48,96+\dfrac{2100}{11}-0,24.233}.100\%=12,786\%\)
