Đáp án:

 a) 44,32g

b) 11,65 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\\
S{O_2} + Ba{(OH)_2} \to BaS{O_3} + {H_2}O\\
n{O_2} = \dfrac{{4,928}}{{22,4}} = 0,22\,mol\\
nS{O_2} = \dfrac{{0,22 \times 8}}{{11}} = 0,16\,mol\\
nBaS{O_3} = nS{O_2} = 0,16\,mol\\
a = mBaS{O_3} = 0,16 \times 217 = 34,72g\\
nFe{S_2} = \dfrac{{0,16}}{2} = 0,08\,mol\\
m = mFe{S_2} = 0,08 \times 120 = 9,6g\\
a + m = 9,6 + 34,72 = 44,32g\\
b)\\
nS{O_3} = \dfrac{8}{{80}} = 0,1\,mol\\
nBa{(OH)_2} = \dfrac{{200 \times 6,4125\% }}{{171}} = 0,075\,mol\\
Ba{(OH)_2} + S{O_3} \to BaS{O_4} + {H_2}O\\
BaS{O_4} + S{O_3} + {H_2}O \to Ba{(HS{O_4})_2}\\
nBaS{O_4} = 0,075 - 0,025 = 0,05\,mol\\
mBaS{O_4} = 0,05 \times 233 = 11,65g
\end{array}\)