Đáp án:

$m_{CaCO_3}=36(g)\\ a=4,818\\ V_{H_2}=5,376(l)\\ C\%_{dd\ FeCl_2}=5,89\%\\ C\%_{dd\ FeCl_3}=1,88\%$

Giải thích các bước giải:

$1/\\ n_{Fe_2O_3}=\dfrac{24}{160}=0,15(mol)\\ Fe_2O_3+3CO\xrightarrow{t^0} 2Fe+3CO_2\\ n_{CO_2}=3.n_{Fe_2O_3}.80\%=0,36(mol)\\ CO_2+Ca(OH)_2\to CaCO_3+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,36(mol)\\ m_{CaCO_3}=0,36.100=36(g)\\ 2/\\ n_{Fe_2O_3\ dư}=0,15.(100\%-80\%)=0,03(mol)\\ n_{Fe}=2.n_{Fe_2O_3}.80\%=0,24(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ Fe_2O_3+^HCl\to 2FeCl_3+3H_2O\\ n_{HCl}=2.n_{Fe}+6.n_{HCl}=0,66(mol)\\ \to C\%_{dd\ HCl}=\dfrac{0,66.36,5}{500}.100\%=4,818\%\\ \to a=4,818\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,24(mol)\\ \to V_{H_2}=0,24.22,4=5,376(l)\\ n_{FeCl_3}=2.n_{Fe_2O_3}=0,06(mol)\\ m_{dd\ B1}=m_{A}+m_{dd\ HCl}-m_{H_2}\\ =m_{Fe}+m_{Fe_2O_3\ dư}+m_{dd\ HCl}-m_{H_2}\\ =0,24.56+0,03.160+500-0,24.2\\ =517,76(g)\\ C\%_{dd\ FeCl_2}=\dfrac{0,24.127}{517,76}.100\%=5,89\%\\ C\%_{dd\ FeCl_3}=\dfrac{0,06.162,5}{517,76}.100\%=1,88\%$