Đáp án:
a) 12,5 g
b) 18,46%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + CaC{O_3} \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\\
mC{H_3}COOH = m{\rm{dd}} \times C\% = 100 \times 15\% = 15g\\
nC{H_3}COOH = \dfrac{m}{M} = \dfrac{{15}}{{60}} = 0,25\,mol\\
nCaC{O_3} = \dfrac{{0,25}}{2} = 0,125\,mol\\
mCaC{O_3} = n \times M = 0,125 \times 100 = 12,5g\\
b)\\
nC{O_2} = nCaC{O_3} = 0,125\,mol\\
m{\rm{dd}}spu = 100 + 12,5 - 0,125 \times 44 = 107g\\
n{(C{H_3}COO)_2}Ca = nCaC{O_3} = 0,125\,mol\\
C\% {(C{H_3}COO)_2}Ca = \dfrac{{0,125 \times 158}}{{107}} \times 100\% = 18,46\%
\end{array}\)
