A + 2HCl \(\rightarrow\)ACl2 +H2
0,2__0,4___ 0,2___0,2 mol
nH2= \(\frac{4,48}{22,4}\)=0,2 mol
\(\rightarrow\)M A= \(\frac{4,8}{0,2}\)=24(Mg)
nHCl= 200.20/100.36,5=1,1>0.4
\(\rightarrow\) HCl dư
mdd sau= 4,8 + 200 - 0,2.2= 204,4g
C% Mgcl2= 0,2.95.100/204,4=9,295%
C%HCl dư=(1,1-0,4).36,5.100/204,4=12,5%