Đáp án:
b, \({V_{{H_2}}} =4,48l\)
c, \({V_{ddC{H_3}COOH}} = \dfrac{{200}}{{1,02}} = 196,1ml\)
Giải thích các bước giải:
\(\begin{array}{l}
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
{n_{Mg}} = 0,2mol\\
\to {n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
\to {V_{{H_2}}} = 4,48l\\
\to {n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,4mol\\
\to {m_{C{H_3}COOH}} = 24g\\
\to {m_{ddC{H_3}COOH}} = \dfrac{{24 \times 100}}{{12}} = 200g\\
\to {V_{ddC{H_3}COOH}} = \dfrac{{200}}{{1,02}} = 196,1ml
\end{array}\)