4Al+3O2--->2Al2O3

2Mg+O2--->2MgO

4Na+O2--->2Na2O

mO2=m chất rắn-mhh=17-10,2=6,8(g)

=>nO2=6,8/32=0,2125(mol)

2Al+6HCl--->2AlCl3+3H2

Mg+2HCl--->MgCl2+H2

2Na+2HCl--->2NaCl+H2

Theo pt: nH2=2nO2=2.0,2125=0,425(mol)

=>VH2=0,425.22,4=9,52(l)

Theo pt: nHCl=2nH2=2.0,425=0,85(mol)

=>nCl-=0,85(mol)

=>mCl-=0,85.35,5=30,175(g)

m muối=m kim loại +mCl-=10,2+30,175=40,375(g)

$n_{O_2}=\dfrac{17-10,22}{32}=0,211875(mol)$

$4Al+3O_2\xrightarrow{{t^o}} 2Al_2O_3$

$2Mg+O_2\xrightarrow{{t^o}} 2MgO$

$4Na+O_2\xrightarrow{{t^o}} 2Na_2O$

$\to n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Mg}+\dfrac{1}{4}n_{Na}$

$\to 4n_{O_2}=3n_{Al}+2n_{Mg}+n_{Na}$

$2Al+6HCl\to 2AlCl_3+3H_2$

$Mg+2HCl\to MgCl_2+H_2$

$2Na+2HCl\to 2NaCl+H_2$

$\to n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}+\dfrac{1}{2}n_{Na}$

$\to 2n_{H_2}=3n_{Al}+2n_{Mg}+n_{Na}$

$\to 2n_{H_2}=4n_{O_2}$

$\to n_{H_2}=0,42375(mol)$

$\to V=0,42375.22,4=9,492l$

Theo PTHH, $n_{HCl}=2n_{H_2}=0,8475(mol)$

BTKL:

$m=10,22+0,8475.36,5-0,42375.2=40,30625g$