Mg + 2HCl -> MgCl2 + H2
nMg=0,05(mol)
Theo PTHH ta có:
nH2=nMg=nMgCl2=0,05(mol)
nHCl=2nMg=0,1(mol)
VH2=22,4.0,05=1,12(lít)
mMgCl2=95.0,05=4,75(g)
mHCl=36,5.0,1=3,65(g)
mdd HCl=3,65:18,25%=20(g)
mdd sau PƯ=20+1,2-0,05.2=21,1(g)
C% dd MgCl2=\(\dfrac{4,75}{21,1}.100\%=22,5\%\)