Đáp án:
m=96,9 gam
Giải thích các bước giải:
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
Al \hfill \\
Mg \hfill \\
Fe \hfill \\
\end{gathered} \right. + \left\{ \begin{gathered}
{H_2}S{O_4} \hfill \\
HCl \hfill \\
\end{gathered} \right.\xrightarrow{{}}muối + {H_2}\)
Ta có:
\({n_{{H_2}(1)}} = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}};{n_{{H_2}(2)}} = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}}\)
\( \to {{\text{n}}_{HCl}} = 2{n_{{H_2}(2)}} = 1{\text{ mol}}\)
BTKL: \({m_{kl}} + {m_{{H_2}S{O_4}}} + {m_{HCl}} = {m_{muối}} + {m_{{H_2}}}\)
\( \to 23 + 0,4.98 + 1.36,5 = m + (0,4 + 0,5).2 \to m = 96,9{\text{ gam}}\)