Mg + 2HCI -> $MgCI_{2} + H_{2}$$ n_{Mg} = 0,05 ( mol)$
Theo PTHHta có :
$ n_{H2} = n_{Mg} = n_{MgCI2} = 0,05mol $
$ n_{HCI} = 2n_{Mg} = 0,1mol$
$ V_{H2} = 22,4.0,05 = 1,12(lít)$
$ m_{MgCI2} = 95.0,05 = 4,75 (g)$
$ m_{HCI} = 36,5 . 0,1 = 3,65(g)$
$ m_{dd HCI}$ = 3,65 : 18,25% = 20(g)
$ m_{dd sau PƯ} = 20+1,2-0,05.2=21,1(g)$
C% dd MgCI2 = $ \frac{4,75}{21,1} .$100% = 22,5%