a)
Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b)
\(n_{H_2}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\)
Gọi số mol Al2O3, Al là x, y (mol)
=> 102x + 27y = 5,7 (1)
PTHH: Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
x------->3x--------->x
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
y---->1,5y--------->0,5y------>1,5y
=> 1,5y = 0,09
=> y = 0,06 (mol)
(1) => x = 0,04 (mol)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,06.27}{5,7}.100\%=28,421\%\\\%m_{Al_2O_3}=\dfrac{102.0,04}{5,7}.100\%=71,579\%\end{matrix}\right.\)
c)
\(n_{H_2SO_4}=3x+1,5y=0,21\left(mol\right)\)
=> \(m_{H_2SO_4}=0,21.98=20,58\left(g\right)\)
=> \(m=\dfrac{20,58.100}{19,6}=105\left(g\right)\)
d) \(m_{dd.sau.pư}=5,7+105-0,09.2=110,52\left(g\right)\)
\(n_{Al_2\left(SO_4\right)_3}=x+0,5y=0,07\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,07.342=23,94\left(g\right)\)
=> \(C\%=\dfrac{23,94}{110,52}.100\%=21,66\%\)
