Giải thích các bước giải:
Bài 3:
\(\begin{array}{l}
NaOH + C{H_3}{\rm{COOH}} \to C{H_3}{\rm{COO}}Na + {H_2}O\\
{m_{NaOH}} = \dfrac{{30 \times 20\% }}{{100\% }} = 6g\\
\to {n_{NaOH}} = 0,15mol\\
\to {n_{C{H_3}{\rm{COO}}H}} = {n_{NaOH}} = 0,15mol\\
\to {C_M} = \dfrac{{0,15}}{{0,5}} = 0,3M\\
N{a_2}C{O_3} + 2C{H_3}{\rm{COOH}} \to C{H_3}{\rm{COO}}Na + {H_2}O + C{O_2}\\
{n_{N{a_2}C{O_3}}} = 0,5 \times 0,2 = 0,1mol\\
{n_{C{H_3}{\rm{COO}}H}} = 2 \times 0,15 = 0,3 > {n_{N{a_2}C{O_3}}}\\
\to {n_{C{H_3}{\rm{COO}}Hdư}}\\
\to {n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,1mol\\
\to {V_{C{O_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)
Bài 4:
\(\begin{array}{l}
{C_2}{H_5}OH + 3{O_2} \to 3{H_2}O + 2C{O_2}\\
{n_{C{O_2}}} = 0,2mol\\
\to {n_{{C_2}{H_5}OH}} = \dfrac{1}{2}{n_{C{O_2}}} = 0,1mol\\
\to {m_{{C_2}{H_5}OH}} = 0,1 \times 46 = 4,6g\\
\end{array}\)
\(\begin{array}{l}
{n_{{O_2}}} = 1,5{n_{C{O_2}}} = 0,3mol\\
\to {V_{{O_2}}} = 6,72l\\
\to {V_{KK}} = 5{V_{{O_2}}} = 33,6l
\end{array}\)
\(\begin{array}{l}
{V_{{C_2}{H_5}OH}} = \dfrac{m}{D} = \dfrac{{4,6}}{{0,8}} = 5,75ml\\
\to {V_{{C_2}{H_5}OH{\rm{dd}}}} = \dfrac{{5,75 \times 100}}{{40}} = 14,375ml
\end{array}\)
5,
\(\begin{array}{l}
{C_2}{H_5}OH + 3{O_2} \to 3{H_2}O + 2C{O_2}\\
{n_{{C_2}{H_5}OH}} = 0,1mol\\
\to {n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,3mol \to {V_{{O_2}}} = 6,72l\\
\to {V_{KK}} = 5{V_{{O_2}}} = 33,6l\\
\to {V_{{C_2}{H_5}OH}} = \dfrac{m}{D} = 5,75ml\\
\to {V_{{C_2}{H_5}OH{\rm{dd}}}} = \dfrac{{5,75 \times 100}}{8} = 71,875ml\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
{n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 0,1mol\\
\to {m_{C{H_3}COOH}} = 6g\\
\to {m_{C{H_3}COOH}}thực= \dfrac{{6 \times 80}}{{100}} = 4,8g
\end{array}\)
