7)
Hợp chất tạo bởi \(C;H;O\) nên có dạng \(C_xH_yO_z\)
\( \to {M_X} = 12x + y + 16z = 88\)
\( \to \% {m_C} = \frac{{12x}}{{{M_X}}} = \frac{{12x}}{{88}} = 54,54\% \to x = 4\)
\(\% {m_H} = \frac{y}{{{M_X}}} = \frac{y}{{88}} = 9,1\% \to y = 8 \to z = 2\)
Vậy \(X\) là \(C_4H_8O_2\)
11)
Sơ đồ phản ứng:
\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)
Ta có:
\({n_{{O_2}}} = \frac{{14,4}}{{32}} = 0,45{\text{ mol}}\)
\({n_{C{O_2}}} = \frac{{13,2}}{{44}} = 0,3{\text{ mol = }}{{\text{n}}_C}\)
\({n_{{H_2}O}} = \frac{{7,2}}{{18}} = 0,4{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,8{\text{ mol}}\)
Bảo toàn \(O\)
\({n_{O{\text{ trong A}}}} + 2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}\)
\( \to {n_{O{\text{ trong A}}}} + 0,45.2 = 0,3.2 + 0,4 \to {n_{O{\text{ trong A}}}} = 0,1{\text{ mol}}\)
\( \to {C_A} = \frac{{{n_C}}}{{{n_A}}} = \frac{{0,3}}{{0,1}} = 3\)
\({H_A} = \frac{{{n_H}}}{{{n_A}}} = \frac{{0,8}}{{0,1}} = 8\)
\({O_A} = \frac{{{n_O}}}{{{n_A}}} = \frac{{0,1}}{{0,1}} = 1\)
Vậy \(A\) là \(C_3H_8O\)
