Gọi mMg = mZn = mFe = a(g)
\(n_{Mg}=\dfrac{a}{24}\left(mol\right),n_{Al}=\dfrac{a}{27}\left(mol\right)\\ n_{Zn}=\dfrac{a}{65}\left(mol\right),n_{Fe}=\dfrac{a}{56}\left(mol\right)\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(\dfrac{a}{24}\) --> \(\dfrac{a}{24}\left(mol\right)\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(\dfrac{a}{27}\) --> \(\dfrac{a}{16}\left(mol\right)\) (2)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(\dfrac{a}{65}\) --> \(\dfrac{a}{65}\left(mol\right)\) (3)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(\dfrac{a}{56}\) --> \(\dfrac{a}{56}\left(mol\right)\) (4)
Từ (1),(2),(3),(4) có: \(\dfrac{a}{16}>\dfrac{a}{24}>\dfrac{a}{56}>\dfrac{a}{65}\)
Vậy \(V_{H_2}\) thoát ra từ kim loại \(Al\) là lớn nhất