$NaOH + HCl \rightarrow NaCl + H_2O$
a
$KOH + HCl \rightarrow KCl + H_2O$
b
Ta có:
$40a+56b=3,240$
$58,5a+74,5b=4,535$
\Rightarrow $a=0,0425$ ; $b=0,0275$
\Rightarrow $\sum n_{HCl}=a+b=0,07$ (mol)
$n_{NaCl}=a=0,0425(mol)$ \Rightarrow $m_{NaCl}=2,486(g)$
\Rightarrow $B$