Đáp án:
\({V_{{H_2}}} = 4,48{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{4,8}}{{24}} = 0,2{\text{ mol = }}{{\text{n}}_{{H_2}}} \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({n_{HCl}} = 2{n_{Mg}} = 0,2.2 = 0,4{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}} \to {{\text{m}}_{dd{\text{HCl}}}} = \frac{{14,6}}{{10\% }} = 146{\text{ gam}}\)
\({n_{MgC{l_2}}} = {n_{Mg}} = 0,2{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,2.(24 + 35,5.2) = 19{\text{ gam}}\)
BTKL:
\({m_{Mg}} + {m_{dd{\text{ HCl}}}} = {m_{dd{\text{ }}muối}} + {m_{{H_2}}} \to {m_{dd{\text{ muối}}}} = 4,8 + 146 - 0,2.2 = 150,4{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{19}}{{150,4}} = 12,633\% \)
