+) \({n_{S{O_2}}} = a{\mkern 1mu} mol;{\mkern 1mu} {n_{{H_2}S}} = b{\mkern 1mu} mol\)

n_X = 4,144 : 22,4= 0,185 mol → a + b = 0,185 (1)

+) \({\bar M_X} = {\kern 1pt} \frac{{64a + 34b}}{{a + b}} = 2.31,595\)  → 0,81a = 29,19b  (2)

Từ (1) và (2) → a = 0,18;  b = 0,005

Xét quá trình nhận e

\(\begin{array}{l} \mathop S\limits^{ + 6} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2e{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \to {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop S\limits^{ + 4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,0,36{\mkern 1mu} {\mkern 1mu} \leftarrow {\mkern 1mu} {\mkern 1mu} 0,18\\ \mathop S\limits^{ + 6} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 8e{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \to {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop S\limits^{ - 2} \\ {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0,04{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \leftarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0,005 \end{array}\)

→ n_{e nhận} = 0,36 + 0,04 = 0,4 mol → n_{e cho}  = 0,4 mol

+) \({n_{{H_2}S{O_4}}} = {n_{SO_4^{2 - }}} + {n_{S{O_2}}} + {n_{{H_2}S}} = \frac{{{n_{e\,cho}}}}{2} + 0,18 + 0,005\) = 0,385 mol 

→mH2SO4 = 0,385.98 = 37,73 gam

Đáp án cần chọn là: A