Đáp án:
\(\begin{array}{l}
14.C\\
15.D\\
16.A\\
17.A\\
18.B\\
19.C\\
20.B
\end{array}\)
Giải thích các bước:
15,
\(Zn + 2HCl \to ZnC{l_2} + {H_2}\)
16,
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 0,1mol\\
{n_{HCl}} = 0,1mol\\
\to {n_{Fe}} > \dfrac{{{n_{HCl}}}}{2}
\end{array}\)
Suy ra Fe dư
\(\begin{array}{l}
{n_{Fe}}dư= 0,1 - 0,05 = 0,05mol\\
\to {m_{Fe}}dư= 2,8g
\end{array}\)
20,
\(Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\)
