Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Na_2CO_3}=\frac{10}{106}=0,094(mol)$

$Na_2CO_3+2CH_3COOH\to 2CH_3COONa+CO_2+H_2O$

Theo PT: $n_{CH_3COOH}=2n_{Na_2CO_3}=0,188(mol)$

$\to V_{dd\,CH_3COOH}=\frac{0,188}{0,5}=0,376(l)=376(ml)$

$b\big)$

Theo PT: $n_{CO_2}=n_{Na_2CO_3}=0,094(mol)$

$n_{Ca(OH)_2}=0,075.1=0,075(mol)$

$\to n_{OH^-}=0,15(mol)$

$\to T=\frac{0,15}{0,094}=1,6$

$\to$ Tạo 2 muối $CaCO_3:x(mol);Ca(HCO_3)_2:y(mol)$

Bảo toàn C: $x+2y=0,094$

Bảo toàn Ca: $x+y=0,075$

$\to x=0,056(mol);y=0,019(mol)$

$\to m_{KT}=m_{CaCO_3}=0,056.100=5,6(g)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
{n_{N{a_2}C{O_3}}} = \dfrac{{10}}{{106}} = 0,094mol\\
N{a_2}C{O_3} + 2C{H_3}COOH \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
{n_{C{H_3}COOH}} = 2{n_{N{a_2}C{O_3}}} = 0,188mol\\
{V_{C{H_3}COOH}} = \dfrac{{0,188}}{{0,5}} = 0,376l = 376ml\\
b)\\
{n_{Ca{{(OH)}_2}}} = 0,075 \times 1 = 0,075mol\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,094mol\\
T = \frac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \frac{{0,075}}{{0,094}} = 0,8\\
 \Rightarrow\text{Phản ứng tạo ra 2 muối } CaC{O_3},Ca{(HC{O_3})_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
Ca{(OH)_2} + 2C{O_2} \to Ca{(HC{O_3})_2}\\
hh:CaC{O_3}(a\,mol),Ca{(HC{O_3})_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,075\\
a + 2b = 0,094
\end{array} \right.\\
 \Rightarrow a = 0,056;b = 0,019\\
{m_{CaC{O_3}}} = 0,056 \times 100 = 5,6g
\end{array}\)