a)
Đặt \(\left\{{}\begin{matrix}n_{CaO}=x\left(mol\right)\\n_{Al_2O_3}=y\left(mol\right)\end{matrix}\right.\)
=> 40x + 102y = 33,9 (1)
\(n_{H_2SO_4}=2.0,45=0,9\left(mol\right)\)
PTHH:
\(CaO+H_2SO_4\rightarrow CaSO_4+H_2O\)
x------>x
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
y----->3y
=> x + 3y = 0,9 (2)
(1),(2) => \(\left\{{}\begin{matrix}x=0,55\\y=\dfrac{7}{60}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{CaO}=0,55.40=22\left(g\right)\\m_{Al_2O_3}=\dfrac{7}{60}.102=11,9\left(g\right)\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}\%m_{CaO}=\dfrac{22}{33,9}.100\%=64,9\%\\\%m_{Al_2O_3}=100\%-64,9\%=35,1\%\end{matrix}\right.\)
