Đáp án:
\({{\text{V}}_{{H_2}}} = 2,24{\text{ lít}}\)
\(C{\% _{MgC{l_2}}} = 9,3\% ;C{\% _{HCl{\text{ dư}}}} = 7,14\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol;}}{{\text{m}}_{HCl}} = 100.14,6\% = 14,6{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{14,6}}{{36,5}} = 0,4{\text{ mol > 2}}{{\text{n}}_{Mg}}\)
Do vậy HCl dư
\({n_{{H_2}}} = {n_{Mg}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({n_{MgC{l_2}}} = {n_{Mg}} = 0,1{\text{ mol;}}{{\text{n}}_{HCl{\text{ dư}}}} = 0,4 - 0,1.2 = 0,2{\text{ mol}}\)
\({m_{MgC{l_2}}} = 0,1.95 = 9,5{\text{ gam;}}{{\text{m}}_{HCl{\text{ dư}}}} = 0,2.36,5 = 7,3{\text{ gam}}\)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} - {m_{{H_2}}} = 2,4 + 100 - 0,1.2 = 102,2{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{9,5}}{{102,2}} = 9,3\% ;C{\% _{HCl{\text{ dư}}}} = \frac{{7,3}}{{102,2}} = 7,14\% \)
