Đáp án:
\(AlC{l_3}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2xAl + yC{l_2}\xrightarrow{{{t^o}}}2A{l_x}C{l_y}\)
BTKL:
\({m_{Al}} + {m_{C{l_2}}} = {m_{A{l_x}C{l_y}}} \to 1,35 + {m_{C{l_2}}} = 6,675 \to {m_{C{l_2}}} = 5,325{\text{ gam}} \to {{\text{n}}_{Al}} = \frac{{1,35}}{{27}} = 0,05{\text{ mol; }}{{\text{n}}_{C{l_2}}} = \frac{{5,325}}{{71}} = 0,075{\text{ mol}}\)
\( \to x:y = {n_{Al}}:2{n_{C{l_2}}} = 0,05:(0,075.2) = 1:3 \to AlC{l_3}\)
\({V_{C{l_2}}} = 0,075.22,4 = 1,68{\text{ lít}}\)