Đáp án:
\({{\text{V}}_{{H_2}}} = 3,36{\text{ lít}}\)
\( \to C{\% _{MgC{l_2}}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = 5,35\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol; }}{{\text{m}}_{HCl}} = 250.98\% = 24,5{\text{gam}} \to {{\text{n}}_{HCl}} = \frac{{24,5}}{{36,5}} > 2{n_{Mg}}\)
do vậy HCl dư.
\( \to {n_{{H_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} - {m_{{H_2}}} = 3,6 + 250 - 0,15.2 = 253,3{\text{ gam}}\)
\({n_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{m}}_{HCl{\text{ phản ứng}}}} = 0,3.36,5 = 10,95{\text{gam}}\)
\( \to {m_{HCl{\text{ dư}}}} = 24,5 - 10,95 = 13,55{\text{gam; }}{{\text{n}}_{MgC{l_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,15.(24 + 35,5.2) = 14,25{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{14,25}}{{253,3}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = \frac{{13,55}}{{253,3}} = 5,35\% \)
