Đáp án:
\({{\text{m}}_{Al}} = 2,7{\text{ gam;}}{{\text{m}}_{A{l_2}{O_3}}} = 5,1{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{n}}_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,1{\text{ mol}} \to {{\text{m}}_{Al}} = 0,1.27 = 2,7{\text{ gam}} \to {{\text{m}}_{A{l_2}{O_3}}} = 7,8 - 2,7 = 5,1{\text{ gam}} \to {{\text{n}}_{A{l_2}{O_3}}} = \frac{{5,1}}{{102}} = 0,05{\text{ mol}}\)
Ta có:
\({n_{HCl}} = 3{n_{Al}} + 6{n_{A{l_2}{O_3}}} = 3.0,1 + 0,05.6 = 0,6{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,6.36,5 = 21,9{\text{ gam}} \to {{\text{m}}_{dd{\text{HCl}}}} = \frac{{21,9}}{{10\% }} = 219{\text{ gam}}\)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = 7,8 + 219 - 0,15.2 = 226,5{\text{ gam}}\)
\({n_{AlC{l_3}}} = 0,1 + 0,05.2 = 0,2{\text{ mol}} \to {{\text{m}}_{AlC{l_3}}} = 0,2.(27.2 + 35,5.3) = 32,1{\text{ gam}} \to {\text{C}}{{\text{\% }}_{AlC{l_3}}} = \frac{{32,1}}{{226,5}} = 14,17\% \)
