Mg + 2HCl \(\rightarrow\)MgCl2 + H2
nH2=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo pTHH ta có:
nMg=nH2=0,2(mol)
mMg=24.0,2=4,8(g)
mCu=11,3-4,8=6,5(g)
b;
Theo pTHH ta có:
2nMg=nHCl=0,4(mol)
VHCl=\(\dfrac{0,4}{0,5}=0,8\left(lít\right)\)