Mg + H2SO4 \(\rightarrow\)MgSO4 + H2 (1)
Fe + H2SO4 \(\rightarrow\)FeSO4 + H2 (2)
Mg + FeSO4 \(\rightarrow\)MgSO4 + Fe (3)
nH2=\(\dfrac{2,016}{22,4}=0,09\left(mol\right)\)\
Đặt nMg=a
nFe(2)=b
nFe(3)=nMg=a
Ta có:a+b=0,09
mFe(3)-mMg=1,68
56a-24a=1,68
32a=1,68
a=0,0525
b=0,0375
mMg=24.0,0525=1,26(g)
mFe(2)=56.0,0375=2,1(g)