a,

Gọi chung 2 kim loại là $R$

$n_{H_2}=\dfrac{2,912}{22,4}=0,13(mol)$

$2R+2H_2O\to 2ROH+H_2$

$\Rightarrow n_R=2n_{H_2}=0,26(mol)$

$\overline{M}_R=\dfrac{7,67}{0,26}=29,5$

$23<29,5<39\Rightarrow $ hai kim loại là $Na$ (x mol), $K$ (y mol)

$\Rightarrow 23x+39y=7,67$                (1)

Mặt khác $0,5x+0,5y=0,13$                   (2)

$(1)(2)\Rightarrow x=0,154375; y=0,105725$

$m_{dd\text{spu}}=7,67+500-0,13.2=507,41g$

$C\%_{NaOH}=\dfrac{0,154375.40.100}{507,41}=1,22\%$

$C\%_{KOH}=\dfrac{0,105725.56.100}{507,41}=1,17\%$

b,

$n_{ROH}=0,26(mol)$

$2ROH+H_2SO_4\to R_2SO_4+2H_2O$

$\Rightarrow$ còn $0,26-0,04.2=0,18(mol)$ $ROH$

$FeCl_3+3ROH\to Fe(OH)_3+3RCl$

$Al_2(SO_4)_3+6ROH\to 2R(OH)_3+3R_2SO_4$

$3n_{FeCl_3}+6n_{Al_2(SO_4)_3}<n_{ROH}$

$\Rightarrow ROH$ dư 

$n_{Fe(OH)_3}=0,024(mol)$

$n_{Al(OH)_3}=0,016.2=0,032(mol)$

$n_{ROH\text{dư}}=0,18-0,024.3-0,032.3=0,012(mol)$

$Al(OH)_3+ROH\to RAlO_2+2H_2O$

$\Rightarrow n_{Al(OH)_3\text{dư}}=0,032-0,012=0,02(mol)$

$\to m=m_{\downarrow}=0,02.78+0,024.107=4,128g$