Đáp án:

$Fe_2(SO_4)_3\\ Fe_2(SO_4)_3.10H_2O\\ V=1,8144$

Giải thích các bước giải:

$m_{CaCO_3}=7,2(g)\\ n_{CaCO_3}=\dfrac{7,2}{100}=0,072(mol)\\ Oxit\ sắt:\ Fe_2O_n\\ M_2O_n+nCO\xrightarrow{ t^0} 2M+nCO_2\\ CO_2+Ca(OH)_2\to CaCO_3+H_2O\\ \xrightarrow{\text{Từ PTHH}}\ n_{CaCO_3}=n_{CO_2}=0,072(mol)\\ \xrightarrow{\text{Từ PTHH}} \ n_{Fe_2O_n}=\dfrac{1}{n}.n_{CO_2}=\dfrac{0,072}{n}(mol)\\ M_{M_2O_n}=4,176:\dfrac{0,072}{n}=58n\\ \Rightarrow 2M+16n=58n\\ \Rightarrow M=21n\\ \Rightarrow \begin{cases}n=\dfrac{8}{3}\\M=56\ (Fe)\end{cases}\\ \to Oxit:\ Fe_3O_4$

$ \xrightarrow{\text{Từ PTHH}}\ n_{Fe}=\dfrac{3}{4}.n_{CO_2}=0,054(mol)\\ Fe+6H_2SO_4\to Fe_2(SO_4)_3+3SO_2+6H_2O\\ \to \text{Muối kim loại}:\ Fe_2(SO_4)_3\\ \text{5,8(g) chất rắn}:\ Fe_2(SO_4)_3.aH_2O\\ \xrightarrow{\text{Từ PTHH}} \ n_{Fe_2(SO_4)_3}=\dfrac{1}{2}.n_{Fe}=0,027(mol)\\ m_{Fe_2(SO_4)_3}=0,027.400=10,8(g)\\ m_{dd\ Fe_2(SO_4)_3}=\dfrac{m_{Fe_2(SO_4)_3}}{C\%}=\dfrac{10,8}{61,069\%}=17,68(g)\\ m_{dd\ sau}=17,68-5,8=11,88(g)\\ m_{Fe_2(SO_4)_3\ sau}=11,88.57,215\%=6,8(g)\\ n_{Fe_2(SO_4)_3\ sau}=\dfrac{6,8}{400}=0,017(mol)\\ n_{Fe_2(SO_4)_3\ mất}=0,027-0,017=0,01(mol)\\ \Rightarrow 0,01=\dfrac{5,8}{400+18a}\\ \Rightarrow a=10\\ \to CTHH:\ Fe_2(SO_4)_3.10H_2O$

$\xrightarrow{\text{Từ PTHH}} \ n_{SO_2}=\dfrac{3}{2}. _{Fe}=0,081(mol)\\ V_{SO_2}=0,081.22,4=1,8144(mol)\\ V=1,8144$