Đáp án:
X là Al
\(C{\% _{AlC{l_3}}} = 19,33\% ;\;{\text{C}}{{\text{\% }}_{HCl}} = 3,45\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2X + 6HCl\xrightarrow{{}}2XC{l_3} + 3{H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{20,16}}{{22,4}} = 0,9{\text{ mol}} \to {{\text{m}}_X} = \frac{2}{3}{n_{{H_2}}} = 0,6{\text{ mol}} \to {{\text{M}}_X} = \frac{{16,2}}{{0,3}} = 27 \to X:Al\)
BTKL: \({m_{dd\;{\text{A}}}} = {m_{Al}} + {m_{dd\;{\text{HCl}}}} - {m_{{H_2}}} = 16,2 + 400 - 0,9.2 = 414,4{\text{ gam}}\)
Ta có:
\({m_{HCl}} = 400.20\% = 80{\text{ gam; }}{{\text{m}}_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 1,8{\text{ mol}} \to {{\text{m}}_{HCl{\text{ dư}}}} = 80 - 1,8.36,5 = 14,3{\text{ gam}}\)
Ta có: \({n_{AlC{l_3}}} = {n_{Al}} = 0,6{\text{ mol}} \to {{\text{m}}_{AlC{l_3}}} = 0,6.(27 + 35,5.3) = 80,1{\text{ gam}}\)
Ta có:
\(C{\% _{AlC{l_3}}} = \frac{{80,1}}{{414,4}} = 19,33\% ;\;{\text{C}}{{\text{\% }}_{HCl}} = \frac{{14.3}}{{414.4}} = 3,45\% \)
