\(\begin{array}{l}
a)\\
2C{H_3}COOH + CaC{O_3} \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\\
nC{O_2} = \dfrac{{0,896}}{{22,4}} = 0,04\,mol \Rightarrow nCaC{O_3} = nC{O_2} = 0,04\,mol\\
\% mCaC{O_3} = \dfrac{{0,04 \times 100}}{{12,5}} \times 100\% = 32\% \\
\% mNaCl = 100 - 32 = 68\% \\
b)\\
nC{H_3}COOH = 2nC{O_2} = 0,08\,mol\\
C\% C{H_3}COOH = \dfrac{{0,08 \times 60}}{{120}} \times 100\% = 4\% \\
c)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
nN{a_2}C{O_3} = nC{O_2} = 0,04\,mol\\
H = 85\% \Rightarrow mN{a_2}C{O_3} = 0,04 \times 85\% \times 106 = 3,604g
\end{array}\)
