a)
m_{Fe}=28.\frac{40}{100}= 11,2 \rightarrow n_{Fe}=\frac{11,2}{56}=0,2 (mol)
m_{Al}=28- 11,2=16,8\Rightarrow n_{Al}=\frac{16,8}{27}=\frac{28}{45}(mol)
PTHH:
Fe + H_{2}SO_{4}\rightarrow FeSO_{4}+ H_{2}
0,2\rightarrow0,2(mol)
2Al+ H_{2}SO_{4}\rightarrow Al_{2}(SO_{4})_{3}+3H_{2}...