Đáp án + Giải thích các bước giải:

 1. $n_A = \dfrac{V}{22,4} = \dfrac{0,35}{22,4} = \dfrac{1}{64}(mol)$

$\Rightarrow M_A = \dfrac{m}{n} = \dfrac{1}{\dfrac{1}{64}} = 64(g/mol)$
Giả sử có $1$ mol $A$
$\Rightarrow m_S = 50\% . 64 = 32(g)$
$\Rightarrow n_S = \dfrac{m}{M}= \dfrac{32}{32} = 1(mol)$

$m_O = 64 - 32 =32(g)$

$\Rightarrow n_S = \dfrac{m}{M}=  \dfrac{32}{16} = 2(mol)$
Vậy $1$ mol $A$ gồm $1$ mol $S$ và $2$ mol $O$
$\Rightarrow A$ là $SO_2$
2. $n_{SO_2} = \dfrac{V}{22,4} = \dfrac{8}{22,4} = \dfrac{5}{14}(mol)$

PTHH:
$2SO_2 + O_2 \overset{t^o,V_2O_5}{\rightleftharpoons} 2SO_3$

$SO_3 + H_2O \longrightarrow H_2SO_4$
Theo PTHH:
$n_{H_2SO_4} = n_{SO_3} = n_{SO_2} = \dfrac{5}{14}(mol)$
$\Rightarrow m_{H_2SO_4} = n . M = \dfrac{5}{14} . (2 . 1 + 32 + 4 . 16) = \dfrac{5}{14} . 98 = 35(g)$

$\Rightarrow m_{SO_3}=  n . M = \dfrac{5}{14} . 80 = \dfrac{200}{7} \approx 28,57(g)$

$m_{H_2SO_4(dd)} = \dfrac{C\% . m_{dd}}{100\%}=  \dfrac{60\% . 85,8}{100\%}= 51,48(g)$
$\Rightarrow m_{H_2SO_4(dd)} = 51,48 + 35 = 86,48(g)$
$m_{ddH_2SO_4} = 85,8 + 28,57 = 114,37(g)$
$\Rightarrow C\%_{H_2SO_4} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{86,48}{114,37} . 100\% \approx 75,61\%$
$\color{red}{\text{#Vexi}}$

\(\begin{array}{l}
1)\\
Oxit:\,SO_n\,(\rm do\,S\,hóa\,trị\,chẵn)\\
M_{SO_n}=32+16n=\frac{16n}{50\%}=32n\\
\to n=2\\
\to Oxit:\,SO_2\\
2)\\
n_A=\frac{8}{22,4}=\frac{5}{14}(mol)\\
2SO_2+O_2\xrightarrow{t^o,V_2O_5}2SO_3\\
SO_3+H_2O\to H_2SO_4\\
Theo\,PT:\,n_{H_2SO_4(pu)}=n_{SO_2}=\frac{5}{14}(mol)\\
m_{H_2SO_4(bđ)}=85,8.60\%=51,48(g)\\
m_{H_2SO_4(sau)}=\frac{5}{14}.98+51,48=84=86,48(g)\\
m_{dd\,H_2SO_4(sau)}=\frac{5}{14}.64+85,8\approx 108,66(g)\\
\to C\%_{H_2SO_4(sau)}=\frac{86,48}{108,66}.100\%\approx 79,59\%
\end{array}\)