Đáp án:
C. 63,65g
Giải thích các bước giải:
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = 3,65mol\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 1,825mol\\
\to {m_{{H_2}}} = 3,65g\\
{m_{HCl}} = 1000 \times 1,19 = 1190g
\end{array}\)
Theo ĐLBTKL, ta có:
\(\begin{array}{l}
{m_{hh}} + {m_{{\rm{dd}}HCl}} = {m_{{\rm{dd}}A}} + {m_{{H_2}}}\\
\to {m_{hh}} = {m_{{\rm{dd}}A}} + {m_{{H_2}}} - {m_{{\rm{dd}}HCl}} = 1250 + 3,65 - 1190\\
\to {m_{hh}} = 63,65g
\end{array}\)
