Đáp án+Giải thích các bước giải:

 @NhuQuynhQ.T gửi ạ

`1)`

`n_{HCl}=0,1.1=0,1(mol)`

`X:K:x(mol);Al:y(mol)`

`39x+27y=10,5(1)`

`2K+2H_2O->2KOH+H_2`

`x->x->x(mol)`

`2Al+2KOH+2H_2O->2KAlO_2+3H_2`

`y->y(mol)`

`->n_{KOH\ du}=x-y(mol)`

`KOH+HCl->KCl+H_2O`

`0,1←0,1(mol)`

`->x-y=0,1(2)`

`(1)(2)->x=0,2;y=0,1`

`\%m_K={0,2.39}/{10,5}.100\%=74,29\%`

`\%m_{Al}=100-74,29=25,71\%`

`2)`

`Z:K:a(mol);Al:b(mol)`

`39a+27b=10,5+9,5=20(3)`

`2K+2H_2O->2KOH+H_2`

`a->a->a(mol)`

`2Al+2KOH+2H_2O->2KAlO_2+3H_2`

`b->b(mol)`

Do HCl vào B tạo tủa ngay nên KOH hết

`KAlO_2+HCl+H_2O->Al(OH)_3+KCl`

`->a=b(4)`

`(3)(4)->a=b={10}/{33}`

`\%m_K={{10}/{33}.39}/{20}.100\%=59,09\%`

`\%m_{Al}=100-59,09=40,91\%`