Bạn tham khảo:
$a/$
BTKL:
$m_{hh}+m_{H_2O}=m_X+m_{H_2}$
$8,5+191,8=200+m_{H_2}$
$\to m_{H_2}=0,3(g)$
$n_{H_2}=\frac{0,3}{2}=0,15(mol)$
$V_{H_2}=0,15.22,4=3,36(l)$
$b/$
Gọi 2 kim là $R$
$2R+2HCl \to 2RCl+H_2$
$\to n_R=0,15.2=0,3(mol)$
$M_R=\frac{8,5}{0,3}=28,33(g/mol)$
$M_{Na}=23 \leq M_R=2833 \leq M_K=39$
$\to R: Na; K$
$c/$
$2Na+2H_2O \to 2NaOH+H_2$
$2K+2H_2O \to 2KOH+H_2$
$n_{Na}=a(mol)$
$n_K=b(mol)$
$m_{hh}=23a+39b=8,5(1)$
$n_{H_2}=0,5a+0,5b=0,15(2)$
$(1)(2)$
$a=0,2$
$b=0,1$
$C\%_{NaOH}=\frac{0,2.40}{200}.100\%=4\%$
$C\%_{KOH}=\frac{0,1.56}{200}.100\%=2,8\%$
