Đáp án:
\({C_2}{H_4}{O_2}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{{O_2}}} = \frac{{5,376}}{{22,4}} = 0,24{\text{ mol}}\)
\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
\(Ca{(OH)_2} + 2C{O_2}\xrightarrow{{}}Ca{(HC{O_3})_2}\)
\({n_{CaC{O_3}}} = \frac{{10}}{{100}} = 0,1{\text{ mol;}}{{\text{n}}_{Ca{{(HC{O_3})}_2}}} = 0,35.0,2 = 0,07{\text{ mol}}\)
Bảo toàn nguyên tố C:
\({n_{C{O_2}}} = {n_{CaC{O_3}}} + 2{n_{Ca{{(HC{O_3})}_2}}} = 0,1 + 0,07.2 = 0,24{\text{ mol}}\)
\({m_{dd{\text{tăng}}}} = {m_{C{O_2}}} + {m_{{H_2}O}} - {m_{CaC{O_3}}} = 4,88{\text{ gam}} \to {\text{0}}{\text{,24}}{\text{.44 + }}{{\text{m}}_{{H_2}O}} - 10 = 4,88{\text{ gam}} \to {{\text{m}}_{{H_2}O}} = 4,32{\text{ gam}} \to {{\text{n}}_{{H_2}O}} = \frac{{4,32}}{{18}} = 0,24{\text{ mol}}\)
Bảo toàn nguyên tố:
\({n_C} = {n_{C{O_2}}} = 0,24{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,48{\text{ mol;}}{{\text{n}}_O} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,24.2 + 0,24 - 0,24.2 = 0,24{\text{ mol}}\)
\( \to {n_C}:{n_H}:{n_O} = 0,24:0,48:0,24 = 1:2:1\)
A có dạng \({(C{H_2}O)_n} \to {M_A} = 30n \to 40 < 30n < 74 \to n = 2 \to A:{C_2}{H_4}{O_2}\)
