2Al + 3H2SO4 -> Al2(SO4)3 + 3 H2 (1)
Mg + H2SO4 ->MgSO4 + H2 (2)
nH2=\(\dfrac{5,04}{22,4}=0,225\left(mol\right)\)
Đặt nAl=a
nMg=b
Ta có hệ:
\(\left\{{}\begin{matrix}27a+24b=4,5\\1,5a+b=0,225\end{matrix}\right.\)
=>a=0,01;b=0,075
mAl=27.0,1=2,7(g)
%mAl=\(\dfrac{2,7}{4,5}.100\%=60\%\)
%mMg=100-60=40%