Đáp án:
m=12,05
Giải thích các bước giải:
\(\begin{array}{l}
C{H_3}COOCH = C{H_2} + KOH \to C{H_3}COOK + C{H_3}CHO\\
{n_{C{H_3}COOCH = C{H_2}}} = \dfrac{{6,45}}{{86}} = 0,075\,mol\\
{n_{KOH}} = \dfrac{{28 \times 20\% }}{{56}} = 0,1\,mol\\
{n_{C{H_3}COOCH = C{H_2}}} < {n_{KOH}} \Rightarrow KOH \text{ dư }\\
{n_{C{H_3}COOK}} = {n_{C{H_3}CHO}} = {n_{C{H_3}COOCH = C{H_2}}} = 0,075\,mol\\
{n_{KOH}} \text{ dư }= 0,1 - 0,075 = 0,025\,mol\\
m = 0,075 \times 98 + 0,075 \times 44 + 0,025 \times 56 = 12,05g
\end{array}\)