Đáp án:

\(\begin{array}{l}
14,A\\
15,D\\
16,B\\
17,A
\end{array}\)

Giải thích các bước giải:

14,

\(\begin{array}{l}
CaC{O_3} \to CaO + C{O_2}\\
C + {O_2} \to C{O_2}\\
{n_{CaO}} = 0,075mol\\
{n_C} = 0,1mol\\
 \to {n_{C{O_2}}} = {n_{CaO}} + {n_C} = 0,175mol\\
 \to {V_{C{O_2}}} = 3,92ml = 3920l
\end{array}\)

15,

\(\begin{array}{l}
{H_x}{S_y}\\
{M_X} = 17 \times 2 = 34\\
{n_H}:{n_S} = x:y = \dfrac{{5,88}}{1}:\dfrac{{94,12}}{{32}} = 5,88:2,94\\
 \to {n_H}:{n_S} = x:y = 2:1\\
 \to {({H_2}S)_n}\\
 \to 34n = 34\\
 \to n = 1\\
 \to {H_2}S\\
 \to 2:1
\end{array}\)

16,

\(\begin{array}{l}
2Mg + {O_2} \to 2MgO\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
{m_{Kim loại}} + {m_{{O_2}}} = {m_{Oxit}}\\
 \to {m_{{O_2}}} = {m_{Oxit}} - {m_{Kim loại}} = a + 2,4 - a = 2,4g\\
 \to {n_{{O_2}}} = 0,075mol\\
 \to V = {V_{{O_2}}} = 1,68l
\end{array}\)

17,

\(\begin{array}{l}
2R + {O_2} \to 2RO\\
{n_R} = {n_{RO}}\\
 \to \dfrac{{7,2}}{R} = \dfrac{{12}}{{R + 16}}\\
 \to R = 24\\
 \to Mg
\end{array}\)