8Al + 15H2SO4 → 4Al2(SO4)3 + 3H2S + 12H2O
\(n_{Al}=\dfrac{1,35}{27}=0,05\left(mol\right)\\ n_{H_2SO_4}=\dfrac{15}{8}n_{Al}=\dfrac{3}{32}\left(mol\right)\\ \Rightarrow m_{ddH_2SO_4}=\dfrac{\dfrac{3}{32}.98}{98\%}=9,375\left(g\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,025\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,025.342=8,55\left(g\right)\\ n_{H_2S}=\dfrac{3}{8}n_{Al}=0,01875\left(mol\right)\\ V_{H_2S}=0,01875.22,4=0,42\left(l\right)=420ml\)
`2Al + 6H_2 SO_4 -> Al_2 (SO_4)_3 + 3SO_2 + 6H_2 O`
`0,05` `0,15` `0,025` `0,075` `(mol)`
`n_[Al]=[1,35]/27=0,05(mol)`
`@ m_[dd H_2 SO_4]=[0,15.98]/98 . 100 = 15 (g)`
`@ m_[Al_2 (SO_4)_3]=0,025.342=8,55(g)`
`@V_[SO_2]=0,075.22,4=1,68(l)`
\(n_{Al}=\dfrac{1,35}{27}=0,05\left(mol\right)\\ pthh:2Al+6H_2SO_4\rightarrow3SO_2+Al_2\left(SO_4\right)_3+6H_2O\)
0,05 0,15 0,225 0,025
\(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\\ m=m_{\text{dd}\left(H_2SO_4\right)}=\dfrac{14,7.100}{98}=15\left(g\right)\\ m_{Al_2\left(SO_4\right)_3}=0,025.342=8,55g\\ V_{SO_2}=0,225.22,4=5,04l\)
=> \(V_{SO_2\left(ml\right)}=5,04.1000=5040ml\)
