Đáp án:
\({\text{C}}{{\text{\% }}_{MgC{l_2}}} = 3,5\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = 2,7\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)
\({n_{MgO}} = \frac{{1,5}}{{40}} = 0,0375{\text{ mol; }}{{\text{m}}_{HCl}} = 100.5,475\% = 5,475{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{5,475}}{{36,5}} = 0,15{\text{ mol > 2}}{{\text{n}}_{MgO}}\) do vậy HCl dư.
\(\to {n_{MgC{l_2}}} = {n_{MgO}} = 0,0375{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,0375.(24 + 35,5.2) = 3,5625{\text{ gam}}\)
\({n_{HCl{\text{ dư}}}} = 0,15 - 2{n_{MgO}} = 0,075{\text{ mol}} \to {{\text{m}}_{HCl{\text{ dư}}}} = 0,075.36,5 = 2,7375{\text{ gam}}\)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = 1,5 + 100 = 101,5{\text{ gam}} \to {\text{C}}{{\text{\% }}_{MgC{l_2}}} = \frac{{3,5625}}{{101,5}} = 3,5\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = \frac{{2,7375}}{{101,5}} = 2,7\% \)
