Đáp án:

1) 15% và 39,75 g

 2) C3H8O

Giải thích các bước giải:

1)

\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
b)\\
nC{O_2} = \dfrac{V}{{22,4}} = \dfrac{{8,4}}{{22,4}} = 0,375\,mol\\
nC{H_3}COOH = 2nC{O_2} = 0,75\,mol\\
nN{a_2}C{O_3} = nC{O_2} = 0,375\,mol\\
a = mN{a_2}C{O_3} = n \times M = 0,375 \times 106 = 39,75g\\
C\% C{H_3}COOH = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\%  = \dfrac{{0,75 \times 60}}{{300}} \times 100\%  = 15\% 
\end{array}\)

 2)

\(\begin{array}{l}
nC{O_2} = \dfrac{m}{M} = \dfrac{{6,6}}{{44}} = 0,15\,mol\\
n{H_2}O = \dfrac{m}{M} = \dfrac{{3,6}}{{18}} = 0,2\,mol\\
nC = nC{O_2} = 0,15\,mol\\
nH = 2n{H_2}O = 0,2 \times 2 = 0,4\,mol\\
mO = 3 - 0,15 \times 12 + 0,4 = 0,8\,g\\
nO = \dfrac{m}{M} = \dfrac{{0,8}}{{16}} = 0,05\,mol\\
nC:nH:nO = 0,15:0,4:0,04 = 3:8:1\\
 \Rightarrow CTDGN:{C_3}{H_8}O\\
MA = 60g/mol \Rightarrow 60n = 60 \Rightarrow n = 1\\
 \Rightarrow CTPT:{C_3}{H_8}O
\end{array}\)

CTCT: \({C_3}{H_7}OH\)