a,

$\Delta m=8-6=2g=m_O$

$CO+O\to CO_2$

$\Rightarrow n_{CO_2}=n_O=\dfrac{2}{16}=0,125(mol)$

$\to V=0,125.22,4=2,8l$

b,

$n_{Ba(OH)_2}=0,1.1,2=0,12(mol)$

$\dfrac{2.0,12}{0,125}=1,92$

$\to$ tạo 2 muối

$Ba(OH)_2+CO_2\to BaCO_3+H_2O$

$BaCO_3+CO_2+H_2O\to Ba(HCO_3)_2$

$\Rightarrow n_{BaCO_3}=n_{Ba(OH)_2}- n_{CO_2\text{dư}}=0,12-(0,125-0,12)=0,115(mol)$

$\to m=0,115.197=22,665g$

Đáp án:

 a) 2,8l

b) 22,655g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 3CO \xrightarrow{t^0} 2Fe + 3C{O_2}\\
CuO + CO \xrightarrow{t^0} Cu + C{O_2}\\
\text{ Gọi a là số mol của $CO_2$}\\
{n_{CO}} = {n_{C{O_2}}} = a\,mol\\
BTKL:\,{m_{hh}} + {m_{CO}} = {m_{hh}} + {m_{C{O_2}}}\\
 \Leftrightarrow 8 + 28a = 6 + 44a \Leftrightarrow a = 0,125\,mol\\
V = 0,125 \times 22,4 = 2,8l\\
b)\\
{n_{Ba{{(OH)}_2}}} = 0,1 \times 1,2 = 0,12\,mol\\
T = \dfrac{{{n_{Ba{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,12}}{{0,125}} = 0,96\\
0,5 < T < 1 \Rightarrow  \text{ Tạo cả 2 muối }\\
Ba{(OH)_2} + C{O_2} \to BaC{O_3} + {H_2}O(1)\\
BaC{O_3} + C{O_2} + {H_2}O \to Ba{(HC{O_3})_2}(2)\\
{n_{C{O_2}(1)}} = {n_{BaC{O_3}(1)}} = {n_{Ba{{(OH)}_2}}} = 0,12\,mol\\
{n_{BaC{O_3}(2)}} = {n_{C{O_2}(2)}} = 0,125 - 0,12 = 0,005\,mol\\
{n_{BaC{O_3}}} = 0,12 - 0,005 = 0,115\,mol\\
{m_{BaC{O_3}}} = 0,115 \times 197 = 22,655g
\end{array}\)