Đáp án:
a) \({\text{\% }}{{\text{m}}_{Al}} = 42,86\% \to \% {m_{Mg}} = 57,14\% \)
b) V=6,72 lít
c) \(C{\% _{AlC{l_3}}} = 5,94\% ;{\text{ C}}{{\text{\% }}_{MgC{l_2}}} = 6,34\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Gọi số mol Al là x; Mg là y
\(\to 27x + 24y = 6,3\)
Ta có: \({m_{HCl}} = 219.10\% = 21,9{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{21,9}}{{36,5}} = 0,6{\text{ mol = }}3{n_{Al}} + 2{n_{Mg}} = 3x + 2y\)
GIải được: x=0,1; y=0,15.
\(\to {m_{Al}} = 0,1.27 = 2,7{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Al}} = \frac{{2,7}}{{6,3}} = 42,86\% \to \% {m_{Mg}} = 57,14\% \)
Ta có: \({n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,3{\text{ mol}} \to {\text{V = 0}}{\text{,3}}{\text{.22}}{\text{,4 = 6}}{\text{,72 lít}}\)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = 6,3 + 219 - 0,3.2 = 224,7{\text{ gam}}\)
\({n_{AlC{l_3}}} = {n_{Al}} = 0,1{\text{ mol}} \to {{\text{m}}_{AlC{l_3}}} = 13,35{\text{ gam}} \to C{\% _{AlC{l_3}}} = \frac{{13,35}}{{224,7}} = 5,94\% \)
\({n_{MgC{l_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 14,25{\text{ gam}} \to {\text{C}}{{\text{\% }}_{MgC{l_2}}} = \frac{{14,25}}{{224,7}} = 6,34\% \)
