Đáp án + Giải thích các bước giải:

$n_{CaCO_3} = \dfrac{15}{100} = 0,15 (mol)$

PTHH: 

$a)Ba(OH)_2 + Ca(HCO_3)_2 \rightarrow BaCO_3 \downarrow + CaCO_3 \downarrow + 2H_2O(1)$

$\rightarrow$ 29,7gam kết tủa gồm: $BaCO_3, CaCO_3$

Theo PT $(1): n_{Ca(HCO_3)_2} = n_{BaCO_3} = n_{CaCO_3} = \dfrac{29,7}{197+100} = 0,1 (mol)$

PTHH: 

$Ca(OH)_2 + 2CO_2 \rightarrow Ca(HCO_3)_2(2)$

$Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O(3)$

Theo PT $(2),(3): n_{CO_2} = 2.0,1 + 0,15 = 0,35 (mol)$

PTHH: $C_6H_{12}O_6 \xrightarrow[enzim, H^+]{t^o} 2C_2H_5OH + 2CO_2(4)$

Theo PT $(4): n_{C_6H_{12}O_6} = \dfrac{1}{2} n_{CO_2} = 0,175 (mol)$

$\rightarrow m = 0,175.180 31,5 (g)$

$b)$ Theo PT $(4): n_{C_2H_5OH} = n_{CO_2} = 0,35 (mol)$

$\rightarrow m_{C_2H_5OH} = 0,35.46 = 16,1 (g)$

$\rightarrow V_{C_2H_5OH} = \dfrac{16,1}{0,8} = 20,125 (ml)$

$\rightarrow V_{ddC_2H_5OH} = \dfrac{20,125.100}{90} = \dfrac{805}{36} (ml)$

\(\begin{array}{l}
a)\\
C_6H_{12}O_6\xrightarrow[enzim]{30-32^oC}2C_2H_5OH+2CO_2(1)\\
CO_2+Ca(OH)_2\to CaCO_3↓+H_2O(2)\\
2CO_2+Ca(OH)_2\to Ca(HCO_3)_2(3)\\
Ca(HCO_3)_2+Ba(OH)_2\to BaCO_3↓+CaCO_3↓+H_2O(4)\\
n_{CaCO_3(2)}=\frac{15}{100}=0,15(mol)
Theo\,PT:\,n_{BaCO_3}=n_{CaCO_3(4)}\\
Mat\,khac\,197n_{BaCO_3}+100n_{CaCO_3}=29,7\\
\to n_{BaCO_3}=n_{CaCO_3}=0,1(mol)\\
Theo\,PT:\,\sum n_{CO_2}=n_{CaCO_3(2)}+2n_{CaCO_3(4)}=0,35(mol)\\
Theo\,PT:\,n_{C_6H_{12}O_6}=\frac{1}{2}n_{CO_2}=0,175(mol)\\
\to m=m_{C_6H{12}O_6}=0,175.180=31,5(g)\\
b)\\
Theo\,PT:\,n_{C_2H_5OH}=n_{CO_2}=0,35(mol)\\
\to m_{C_2H_5OH}=0,35.46=16,1(g)\\
\to V_{C_2H_5OH}=\frac{16,1}{0,8}=20,125(ml)\\
\to V_{dd\,rượu}=\frac{20,125.100}{90}≈22,361(ml)
\end{array}\)