$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$

$b,n_{Mg}=\dfrac{4,8}{24}=0,2mol.$

$Theo$ $pt:$ $n_{H_2}=n_{Mg}=0,2mol.$

$⇒V_{H_2}=0,2.22,4=4,48l.$

$c,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$

$Theo$ $pt:$ $n_{Na}=2n_{H_2}=0,4mol.$

$⇒m_{Na}=0,4.23=9,2g.$

$d,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$

$Theo$ $pt:$ $n_{HCl}=2n_{Mg}=0,4mol.$

$⇒m_{HCl}=0,4.36,5=14,6g.$

$⇒m_{ddHCl}=\dfrac{14,6}{7,3\%}=200g.$

Vì axit lấy dư 10% so với lí thuyết nên:

$⇒m_{ddHCl}=200.(100\%+10\%)=220g.$

chúc bạn học tốt!

$n_{Mg}=4,8/24=0,2mol$

$a.Mg+2HCl\to MgCl_2+H_2↑(1)$

b.Theo pt (1) :

$n_{H_2}=n_{Mg}=0,2mol$

$⇒V_{H_2}=0,2.22,4=4,48l$

$c.2Na+2H_2O\to 2NaOH+H_2↑(2)$

Theo pt (2) :

$n_{Na}=2.n_{H_2}=2.0,2=0,4mol$

$⇒m_{Na}=0,4.23=9,2g$

d.Theo pt (1) :

$n_{HCl}=2.n_{Mg}=2.0,2=0,4mol$

$⇒m_{HCl}=0,4.36,5=14,6g$

Vì lượng axit lấy dư 10%

$⇒m_{HCl\ tt}=14,6+14,6.10\%=16,06g$

$⇒m_{dd\ HCl}=\dfrac{16,06}{7,3\%}=220g$