Đáp án:
\({C_7}{H_{12}}{O_5}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(X + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có: \({n_{{O_2}}} = \frac{{1,68}}{{22,4}} = 0,075{\text{ mol}}\)
BTKL:
\({m_{C{O_2}}} + {m_{{H_2}O}} = {m_X} + {m_{{O_2}}} = 1,76 + 0,075.32 = 4,16{\text{gam}}\)
Vì \({V_{C{O_2}}}:{V_{{H_2}O}} = 7:6 \to {V_{C{O_2}}}:{V_{{H_2}O}} = 7:6\)
Gọi \({n_{{H_2}O}} = x \to {n_{C{O_2}}} = \frac{7}{6}x\)
\( \to 44.\frac{7}{6}x + 18.x = 4,16 \to x = 0,06 \to {n_{C{O_2}}} = 0,07;{n_{{H_2}O}} = 0,06{\text{ mol}}\)
\( \to {n_C} = {n_{C{O_2}}} = 0,07;{\text{ }}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,06.2 = 0,12{\text{ mol}} \to {{\text{n}}_O} = \frac{{1,76 - 0,07.12 - 0,12.1}}{{16}} = 0,05{\text{ mol}}\)
\( \to {n_C}:{n_H}:{n_O} = 0,07:0,12:0,05 = 7:12:5\)
A có dạng \({({C_7}{H_{12}}{O_5})_n}\)
\( \to {M_A} = (12.7 + 12 + 16.5)n = 176n < 200 \to n = 1\)
Vậy A là \({C_7}{H_{12}}{O_5}\)
