Đáp án:
\({m_{Fe{\text{ dư}}}} = 2,5315{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol}}\)
\({m_{HCl}} = 20.20\% = 4{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{4}{{36,5}} < 2{n_{Fe}}\)
Do vậy \(Fe\) dư.
\( \to {n_{Fe{\text{ phản ứng}}}} = {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = \frac{2}{{36,5}}{\text{ mol}}\)
\( \to {m_{Fe{\text{ phản ứng}}}} = \frac{2}{{36,5}}.56 = 3,0685{\text{ gam}}\)
\( \to {m_{Fe{\text{ dư}}}} = 5,6 - 3,0685 = 2,5315{\text{ gam}}\)
BTKL:
\({m_{Fe{\text{phản ứng}}}} + {m_{dd{\text{HCl}}}} = {m_{sau{\text{ phản ứng}}}} + {m_{{H_2}}}\)
\( \to 3,0685 + 20 = {m_{dd}} + \frac{2}{{36,5}}.2 \to {m_{dd}} = 22,959{\text{ gam}}\)
\({m_{FeC{l_2}}} = \frac{2}{{36,5}}.(56 + 35,5.2) = 6,959{\text{ gam}}\)
\( \to C{\% _{FeC{l_2}}} = \frac{{6,959}}{{22,959}} = 30,31\% \)
