$n_{Mg}=3,6/24=0,15mol$
$m_{HCl}=259.9,8\%=24,5g⇒n_{HCl}=24,5/36,5=\dfrac{49}{73}mol$
$a/PTHH :$
$Mg+2HCl\to MgCl_2+H_2$
$\text{Theo pt : 1 mol 2 mol}$
$\text{Theo đbài : 0,15mol $\dfrac{49}{73}$}$
$\text{⇒Sau pư HCl dư }$
$\text{Theo pt :}$
$n_{H_2}=n_{Mg}=0,15mol⇒V_{H_2}=0,15.22,4=3,36l$
$\text{Theo ĐLBTKL :}$
$m_{\text{sau pư}}=m_{Mg}+m_{ddHCl}-m_{H_2}=253,3g$
$n_{HCl(pư)}=2.n_{H_2}=2.0,15=0,3mol$
$⇒m_{HCl(pư)}=0,3.36,5=10,95g$
$⇒m_{HCl(dư)}=24,5-10,95=13,55g$
$n_{MgCl_2}=n_{Mg}=0,15mol$
$⇒m_{MgCl_2}=0,15.95=14,25g$
$⇒C\%MgCl_2=\dfrac{14,25}{253,3}.100\%=5,626\%$
$C\%HCl dư=\dfrac{13,55}{253,3}.100\%=5,35\%$
