6)

Ta có:

\({n_{C{O_2}}} = \frac{{3,3}}{{44}} = 0,075{\text{ mol < }}{{\text{n}}_{{H_2}O}} = \frac{{1,8}}{{18}} = 0,1{\text{ mol}}\)

Vậy ancol \(X\) no đơn chức

\( \to {n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,1 - 0,075 = 0,025{\text{ mol}}\)

\( \to {C_X} = \frac{{{n_{C{O_2}}}}}{{{n_X}}} = \frac{{0,075}}{{0,025}} = 3\)

Vậy \(X\) là \(C_3H_8O\)

\( \to {m_X} = 0,025.(12.3 + 8 + 16) = 1,5{\text{ gam}}\)

Chọn \(D\)

7)

Ta có:

\({n_{C{O_2}}} = \frac{{15,4}}{{44}} = 0,35{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{9}{{18}} = 0,5{\text{ mol}}\)

\( \to {n_{hh}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,5 - 0,35 = 0,15{\text{ mol}}\)

Ta có:

\({n_C} = {n_{C{O_2}}} = 0,35{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 1{\text{ mol;}}{{\text{n}}_O} = {n_{hh}} = 0,15{\text{ mol}}\)

\( \to a = {m_C} + {m_H} + {m_O} = 0,35.12 + 1.1 + 0,15.16 = 7,6{\text{ gam}}\)

Chọn \(C\)

8)

Ta có:

\({n_{C{O_2}}} = \frac{{4,704}}{{22,4}} = 0,21{\text{ mol;}}{{\text{n}}_{{H_2}O}} = \frac{{6,12}}{{18}} = 0,34{\text{ mol > }}{{\text{n}}_{C{O_2}}}\)

\( \to {n_{ancol}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,34 - 0,21 = 0,13{\text{ mol}}\)

\( \to {n_C} = {n_{C{O_2}}} = 0,21{\text{ mol;}}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,68{\text{ mol;}}{{\text{n}}_O} = {n_{ancol}} = 0,13{\text{ mol}}\)

\( \to m = {m_C} + {m_H} + {m_O} = 0,21.12 + 0,68.1 + 0,13.16 = 5,28{\text{ gam}}\)

Chọn \(C\)