Câu 1.

Ta có:

\({M_Y} = 3,04.29 = 88 \\\to {n_Y} = \dfrac{{0,88}}{{88}} = 0,01{\text{ mol}}\)

Sơ đồ phản ứng:

\(Y + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \dfrac{{1,76}}{{44}} = 0,04{\text{ mol = }}{{\text{n}}_C};{n_{{H_2}O}} = \dfrac{{0,72}}{{18}} = 0,04{\text{ mol}} \\\to {{\text{n}}_H} = 0,08{\text{ mol}}\)

\( \to {n_O} = \dfrac{{0,88 - 0,04.12 - 0,08.1}}{{16}} = 0,02{\text{ mol}}\)

\( \to {C_Y} = \dfrac{{0,04}}{{0,01}} = 4;{H_Y} = \dfrac{{0,08}}{{0,01}} = 8;\\{O_Y} = \dfrac{{0,02}}{{0,01}} = 2\)

Vậy \(Y\) là \(C_4H_8O_2\)

Câu 2.

Sơ đồ phản ứng:

\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O + {N_2}\)

Áp dụng \(pV=nRT\)

Ta có:

\({n_A} = \dfrac{{0,4.1,64}}{{0,082.(273 + 127)}} = 0,02{\text{ mol}}\)

\({n_{{N_2}}} = \dfrac{{0,112.2}}{{0,082.273}} = 0,01{\text{ mol}}\)

\( \to {n_N} = 2{n_{{N_2}}} = 0,02{\text{ mol}}\)

\({n_{C{O_2}}} = \dfrac{{1,76}}{{44}} = 0,04{\text{ mol = }}\\{{\text{n}}_C};{n_{{H_2}O}} = \dfrac{{0,9}}{{18}} = 0,05{\text{ mol}} \\\to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,1{\text{ mol}}\)

\( \to {n_O} = \dfrac{{1,5 - 0,04.12 - 0,1.1 - 0,02.14}}{{16}} = 0,04{\text{ mol}}\)

\( \to {C_A} = \dfrac{{0,04}}{{0,02}} = 2;\\{H_A} = \dfrac{{0,1}}{{0,02}} = 5;\\{N_A} = \dfrac{{0,02}}{{0,02}} = 1;\\{O_A} = \dfrac{{0,04}}{{0,02}} = 2\)

Vậy \(A\) là \(C_2H_5O_2N\)