PT :\(\left(1\right)NaOH+HCl\rightarrow NaCl+H_2O\)
x mol \(\rightarrow\) x mol
\(\left(2\right)KOH+HCl\rightarrow KCl+H_2O\)
y mol \(\rightarrow\) y mol
Mà ta có :40x + 57y = 15,2
58,5x + 75,5y = 20,75
Giải hệ phương trình có :x=\(\dfrac{19}{170}mol;y=\dfrac{16}{85}mol\)
\(m_{NaOH}=\dfrac{19}{170}.40=4,47g\)
\(\%m_{NaOH}=\dfrac{4,47}{15,2}.100=29,4\%\).
\(\Rightarrow\%m_{KOH}=70,6\%\)
b,
Ta có: \(m_{HCl}=36,5.\dfrac{19}{170}+36,5.\dfrac{16}{85}=43,48g\)
\(m_{dd_{NaOH}}=\dfrac{43,48.100}{7,3}=595,6\%\).
Vậy \(C_{\%_{ddthu}}=\dfrac{20,75}{595,6}.100=3,48\%\) .
