$a)M_X=27.M_{H_2}=27.2=54(g/mol)$

Đặt $n_{SO_2}=x(mol);n_{CO_2}=y(mol)$

$\Rightarrow M_X=\dfrac{64x+44y}{x+y}=54$

$\Rightarrow 64x+44y=54x+54y$

$\Rightarrow 10x=10y\Rightarrow x=y$

$\Rightarrow \%m_{SO_2}=\dfrac{64x}{64x+44y}.100\%=\dfrac{64x}{64x+44x}.100\%=59,26\%$

$\Rightarrow \%m_{CO_2}=100\%-59,26\%=40,74\%$

$b)PTHH:SO_2+Ca(OH)_2\to CaSO_3\downarrow+H_2O$

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

$n_{CO_2}=n_{SO_2}\Rightarrow V_{SO_2}=V_{CO_2}=\dfrac{1}{2}V_X=\dfrac{1}{2}.8,96=4,48(l)$

$\Rightarrow n_{CO_2}=n_{SO_2}=\dfrac{4,48}{22,4}=0,2(mol)$

Theo các PT ta có: $n_{CaCO_3}=n_{CO_2}=n_{SO_2}=n_{CaSO_3}=0,2(mol)$

$\Rightarrow m_{CaSO_3}=0,2.120=24(g);m_{CaCO_3}=0,2.100=20(g)$

Vậy khối lượng kết tủa là $m=m_{CaSO_3}=m_{CaCO_3}=20+24=44(g)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{M_{hh}} = 27 \times 2 = 54g/mol\\
\text{ Giả sử số mol hỗn hợp là 1 mol } \\
hh:C{O_2}(a\,mol),S{O_2}(b\,mol)\\
\left\{ \begin{array}{l}
44a + 64b = 54\\
a + b = 1
\end{array} \right.\\
 \Rightarrow a = b = 0,5\\
\% {m_{C{O_2}}} = \dfrac{{0,5 \times 44}}{{54}} \times 100\%  = 40,74\% \\
\% {m_{S{O_2}}} = 100 - 40,74 = 59,26\% \\
b)\\
{n_{hh}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{S{O_2}}} = {n_{C{O_2}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
Ca{(OH)_2} + S{O_2} \to CaS{O_3} + {H_2}O\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
{n_{CaS{O_3}}} = {n_{S{O_2}}} = 0,2\,mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,2\,mol\\
m = 0,2 \times 120 + 0,2 \times 100 = 44g
\end{array}\)