Em tham khảo nha:
\(\begin{array}{l}
NaOH + HCl \to NaCl + {H_2}O\\
nNaOH = \dfrac{{400 \times 30\% }}{{40}} = 3\,mol\\
nNaCl = nHCl = nNaOH = 3\,mol\\
mNaCl = 3 \times 58,5 = 175,5g\\
mHCl = 3 \times 36,5 = 109,5g\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
nZn = n{H_2} = 0,15\,mol\\
m = 0,15 \times 65 = 9,75g\\
nHCl = 2n{H_2} = 0,3\,mol\\
m{\rm{dd}}HCl = \dfrac{{0,3 \times 36,5}}{{20\% }} = 54,75g
\end{array}\)