Đáp án:
a) 188g
b) 1,35%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
mN{a_2}C{O_3} = m{\rm{dd}} \times C\% = 100 \times 5\% = 5g\\
\Rightarrow nN{a_2}C{O_3} = \dfrac{5}{{106}} \approx 0,047\,mol\\
nC{H_3}COOH = 2N{a_2}C{O_3} = 0,094\,mol\\
m{\rm{dd}}C{H_3}COOH = \dfrac{{0,094 \times 60}}{{3\% }} = 188g\\
b)\\
m{\rm{dd}}spu = 188 + 100 - 0,047 \times 44 = 285,932g\\
C\% C{H_3}COONa = \dfrac{{0,047 \times 82}}{{285,932}} \times 100\% = 1,35\%
\end{array}\)