Đáp án:

 b) 6%

c) 2g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2C{H_3}COOH + CaC{O_3} \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\\
b)\\
nC{O_2} = \dfrac{{0,448}}{{22,4}} = 0,02\,mol\\
nC{H_3}COOH = 2nC{O_2} = 0,04\,mol\\
C\% C{H_3}COOH = \dfrac{{0,04 \times 60}}{{40}} \times 100\%  = 6\% \\
c)\\
nCaC{O_3} = nC{O_2} = 0,02\,mol\\
 \Rightarrow mCaC{O_3} = 0,02 \times 100 = 2g
\end{array}\)

$a/PTHH :$

$2CH_3COOH+CaCO_3\to (CH_3COO)_2Ca+CO_2+H_2O$

$b/n_{CO_2}=0,448/22,4=0,02mol$

$\text{Theo pt :}$

$n_{CH_3COOH}=2.n_{CO_2}=2.0,02=0,04mol$

$⇒m_{CH_3COOH}=0,04.60=2,4g$

$C\%CH_3COOH=\dfrac{2,4.100\%}{40}=6\%$

$\text{c/Theo pt :}$

$n_{CaCO_3}=n_{CO_2}=0,02mol$

$⇒m_{CaCO_3}=0,02.100=2g$