1)

Ta có:

\({n_{C{O_2}}} = \frac{{35,2}}{{44}} = 0,8{\text{ mol}}\)

\( \to {n_{{H_2}O}} = {n_{C{O_2}}} + {n_X} = 0,8 + 0,2 = 1{\text{ mol}}\)

\( \to m = {m_{{H_2}O}} = 1.18 = 18{\text{ gam}}\)

Chọn \(D\)

2)

Sơ đồ phản ứng:

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

\({n_{C{O_2}}} = \dfrac{{1,76}}{{44}} = 0,04{\text{ mol = }}{{\text{n}}_C}\)

\({n_{{H_2}O}} = \dfrac{{0,9}}{{18}} = 0,05{\text{ mol}}\\ \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,1{\text{ mol}}\)

\( \to m = {m_C} + {m_H} = 0,04.12 + 0,1.1 = 0,58{\text{ gam}}\)

Chọn \(D\)

3)

Ta có:

\({n_A} = \dfrac{{6,72}}{{22,4}} = 0,3{\text{ mol;}}\\{{\text{n}}_{{H_2}O}} = \dfrac{{27}}{{18}} = 1,5{\text{ mol}}\)

\( \to {n_{C{O_2}}} = {n_{{H_2}O}} - {n_A} = 1,5 - 0,3 = 1,2{\text{ mol}}\)

\( \to V = {V_{C{O_2}}} = 1,2.22,4 = 26,88{\text{ lít}}\)

4)
Sơ đồ phản ứng:

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

\({n_{C{O_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5{\text{ mol = }}{{\text{n}}_C}\)

\({n_{{H_2}O}} = \dfrac{{10,8}}{{18}} = 0,6{\text{ mol}} \\\to {{\text{n}}_H} = 2{n_{{H_2}O}} = 1,2{\text{ mol}}\)

\( \to m = {m_C} + {m_H} = 0,5.12 + 1,2.1 = 7,2 {\text{ gam}}\)

Chọn \(C\)