1.

a,

Giả sử có x mol K, y mol Al.

$2K+2H_2O\to 2KOH+H_2$  (1)

$\Rightarrow n_{KOH}=x (mol)$

$2Al+2KOH+2H_2O\to 2KAlO_2+3H_2$(2)

Ta thấy khi $y\le x$ thì Al tan hết.

$\Rightarrow \dfrac{n_K}{n_{Al}}=\dfrac{x}{y}\ge 1$

b,

$n_{H_2}=0,525(mol)$

Gọi x, y là số mol K, Al.

$\Rightarrow 39x+27y=18,45$    (*)

Al hết nên ta có: $x+3y=0,525.2$ (bảo toàn e)  (**)

$(*)(**)\Rightarrow x=0,3; y=0,25$

$\Rightarrow m_K=39x=11,7g; m_{Al}=27y=6,75g$

2.

$2K+2HCl\to 2KCl+H_2$

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

$n_{HCl}=0,6.2=1,2(mol)$

$n_{HCl\text{pứ}}=0,3+0,25.3=1,05(mol)$

$\Rightarrow n_{HCl\text{dư}}=0,15(mol)$

$NaOH+HCl\to NaCl+H_2O$

$\Rightarrow n_{NaOH\text{trung hoà}}=0,15(mol)$

$n_{AlCl_3}=0,25(mol)$

$n_{Al(OH)_3}=0,2(mol)<0,25$

- TH1: dư $AlCl_3$

$AlCl_3+3NaOH\to Al(OH)_3+3NaCl$

$\Rightarrow n_{NaOH}=0,2.3=0,6(mol)$

$\Rightarrow C_{M_{NaOH}}=\dfrac{0,6+0,15}{0,4}=1,875M$

- TH2: tủa tan

$AlCl_3+3NaOH\to Al(OH)_3+3NaCl$ (1)

$Al(OH)_3+NaOH\to NaAlO_2+2H_2O$ (2)    

$n_{NaOH(1)}=3n_{AlCl_3}=0,75(mol)$

$n_{Al(OH)_3(1)}=0,25(mol)$ 

$\Rightarrow n_{NaOH(2)}=n_{Al(OH)_3(2)}=0,25-0,2=0,05(mol)$

$\Rightarrow C_{M_{NaOH}}=\dfrac{0,15+0,75+0,05}{0,4}=2,375M$