Đáp án:
6,152g
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,344}}{{22,4}} = 0,06\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,12\,mol\\
BTKL:\\
{m_{hh}} + {m_{HCl}} = a + {m_{{H_2}}} \Leftrightarrow a = {m_{hh}} + {m_{HCl}} - {m_{{H_2}}}\\
\Rightarrow a = 1,892 + 0,12 \times 36,5 - 0,06 \times 2 = 6,152g
\end{array}\)