$a,\ n_{CaCO_3}=\dfrac{16}{100}=0,16(mol)\\ \text{Quy 15(g) hỗn hợp thành}\ Fe\ và\ Fe_2O_3\\ Fe_2O_3+3CO\xrightarrow{t^0} 2Fe+3CO_2\ (1)\\ CO_2+Ca(OH)_2\to CaCO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,16(mol)\\ n_{Fe_2O_3}=\dfrac{1}{3}.n_{CO_2}=\dfrac{0,16}{3}(mol)\\ \Rightarrow n_{Fe\ (1)}=\dfrac{2}{3}.n_{CO_2}=\dfrac{0,32}{3}mol)\\ m_{Fe_2O_3}=\dfrac{0,16}{3}.160=\dfrac{128}{15}(g)\\ \Rightarrow m_{Fe\ BĐ}=15-\dfrac{128}{15}=\dfrac{97}{15}(g)\\ \Rightarrow n_{Fe\ BĐ}=\dfrac{97}{15}:56=\dfrac{97}{840}(mol)\\ \Rightarrow \sum\limits n_{Fe}=\dfrac{0,32}{3}+\dfrac{97}{840}=\dfrac{311}{1400}(mol)\\ \Rightarrow m_{Fe}=\dfrac{311}{1400}.56=12,44(g)\\ b,\ 2Fe+6H_2SO_4\to Fe_2(SO_4)_3+6H_2O+3SO_2\uparrow\\ n_{SO_2}=\dfrac{3}{2}n_{Fe}=\dfrac{311}{1400}.\dfrac{3}{2}\\=\dfrac{933}{2800}(mol)\\ \Rightarrow V_{SO_2}=\dfrac{933}{2800}.22,4=7,464(l)$