$n_{HCl}$=2.1=2(mol)
⇒$m_{HCl}$=2.36,5=73(g)
$n_{H_{2}SO_{4}}$ =1.0,56=0,56(mol)
⇒$m_{H_{2}SO_{4}}$=0,56.98=54,88(g)
$n_{H_{2}}$=$\frac{17,472}{22,4}$ =0,78(mol)
⇒$m_{H_{2}}$=0,78.2=1,56(g)
Phương trình hóa học:
$Mg{}$ + $H_{2}$ $SO_{4}$ → $MgSO_{4}$+$H_{2}$ ↑
$2Al{}$ + $3H_{2}$ $SO_{4}$ → $Al_{2}$$(SO_{4}$)$_{3}$ + $3H_{2}$↑
$Mg{}$ + $2HCl_{}$ → $MgCl_{2}$+ $H_{2}$ ↑
$4Al{}$ + $6HCl_{}$ → $2Al_{2}$$Cl_{3}$ + $3H_{2}$↑
Theo ĐLBTKL ta có :
$m_{kimloại}$ +$m_{axit}$ = $m_{muối}$ +$m_{khí}$
⇒ 15,48 + (73+54,88) = $m_{muối}$+1,56
⇒143,36 =$m_{muối}$+1,56
⇒$m_{muối}$=143,36 -1,56
⇒$m_{muối}$=141,8(g)
@cuthilien
