\(n_{Al_2O_3}=\dfrac{7,65}{102}=0,075mol\)
\(n_{H_2SO_4}=\dfrac{5,88}{98}=0,06mol\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Xét: \(\dfrac{0,075}{1}\) < \(\dfrac{0,06}{3}\) ( mol )
0,02 0,06 0,02 ( mol )
Chất dư là Al2O3
\(m_{Al_2O_3\left(dư\right)}=\left(0,075-0,02\right).102=5,61g\)
\(m_{Al_2\left(SO_4\right)_3}=0,02.342=6,84g\)