\(\text{mcr giảm =16x30%=4,8(g)}\)
\(\Rightarrow\text{m cr giảm=mO trong Fe2O3=4,8(g)}\)
\(\Rightarrow\text{nO=0,3(mol)}\)
\(\Rightarrow\text{nFe=11,2/56=0,2(mol)}\)
=>CTHH là Fe2O3
\(\text{b) Fe2O3+3CO}\rightarrow\text{2Fe+3CO2}\)
\(\text{nCO2=0,3(mol)}\)
\(\text{nCa(OH)2=0,2(mol)}\)
=>Tạo 2 muối
Ca(OH)2(a mol)+CO2->CaCO3+H2O
Ca(OH)2(b mol)+2CO2->Ca(HCO3)2
Ta có:
\(\left\{{}\begin{matrix}\text{a+b=0,2}\\\text{a+2b=0,3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\text{a= 0,1}\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\text{mCa(HCO3)2=16,2(g)}\)
