a) Gọi hh là X :
2X + 2H2O → 2XOH + H2
0,4 0,2 (mol)
\(\overline{X}=\dfrac{11,2}{0,4}=28\)
=> A, B là Na và K
b) \(\left\{{}\begin{matrix}n_{Na}+n_K=0,4\\23.n_{Na}+39.n_K=11,2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n_{Na}=0,275\left(mol\right)\\n_K=0,125\left(mol\right)\end{matrix}\right.\)
2Na + 2H2O → 2NaOH + H2
0,275 0,275
2K + 2H2O → 2KOH + H2
0,125 0,125
200ml≃200g
mdd=11,2+200-0,4.2=208,4g
\(C_{\%\left(NaOH\right)}=\dfrac{0,275.\left(23+16+1\right)}{208,4}.100\simeq5,18\%\)
\(C_{\%\left(KOH\right)}=\dfrac{0,125.\left(39+16+1\right)}{208,4}.100\simeq3,36\%\)
c) 2NaOH + H2SO4 \(\rightarrow\) Na2SO4 + 2H2O
0,275 0,1375
2KOH + H2SO4 \(\rightarrow\) K2SO4 + 2H2O
0,125 0,0625
\(V=\dfrac{n}{C_M}=\dfrac{0,1375+0,0625}{1}=0,2l=200ml\)