Đáp án:
b) 7,5%
c) 9,87%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
b)\\
nC{O_2} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
nC{H_3}COOH = 2nC{O_2} = 0,1\,mol\\
C\% C{H_3}COOH = \dfrac{{0,1 \times 60}}{{80}} \times 100\% = 7,5\% \\
c)\\
nN{a_2}C{O_3} = nC{O_2} = 0,05\,mol\\
m{\rm{dd}}\,spu = 80 + 0,05 \times 106 - 0,05 \times 44 = 83,1g\\
C\% C{H_3}COONa = \dfrac{{0,1 \times 82}}{{83,1}} \times 100\% = 9,87\%
\end{array}\)