a) \(\left\{{}\begin{matrix}n_{Al}=\dfrac{8,64}{27}=0,32\left(mol\right)\\n_{H_2SO_4}=\dfrac{62,72}{98}=0,64\left(mol\right)\end{matrix}\right.\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) Xét tỉ lệ: \(\dfrac{0,32}{2}< \dfrac{0,64}{3}\Rightarrow H_2SO_4\) dư
Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,16\left(mol\right)\\n_{H_2SO_4\left(p\text{ư}\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,48\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{H_2SO_4\left(d\text{ư}\right)}=\left(0,64-0,48\right).96=15,68\left(g\right)\)
c) \(m_{Al_2\left(SO_4\right)_3}=0,16.342=54,72\left(g\right)\)
d) \(V_{H_2}=0,48.22,4=10,752\left(l\right)\)
