nH2=6,72.22,4=0.3mol
a)PTHH : CuO+2HCl->CuCl2+h2o
0,16-->0,32
Mg+2HCl->MgCl2+H2
0,3 <-0.6<----------0.3
b)mMg=0,3.24=7.2g
%mMg=7,2/20 .100%=36%
%mCuO=100%-36%=64%
c) mCuO=mhh-mMg=20-7,2=12.8g
nCuO=12,8/80=0,16mol
Tổng nHCl=0,6+0,32=0,92mol=> mHCl=0,92.36,5=33.58g
mdd sau phản ứng: 20+400-(0,3.2)=419.4g
C%hcl=33.58/419,4 .100%=8%