$n_{Mg}$= $\frac{2}{24}$ =0.083 (mol)
a, PTHH: 2HCl+Mg→H2+MgCl2
0,166<-0.083->0,083 (mol)
=> $V_{H2}$ = 0,083.22,4= 1,86 (l)
$m_{H2}$=0,083.2= 0,166 (g)
b, $m_{HCl(trong dd)}$ = 0,166.(35,5+1)= 6,059 (g)
C%=$\frac{6,059}{200}$ .100% = 3,03%
c, m dd muối = 2+200-0,166=201,834 (g)
m MgCl2 = 0,083. (24+35,5.2)=7,885 (g)
C%=$\frac{7,885}{201,834}$.100%=3,906%