a) \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(0,3>\dfrac{0,2}{3}\Rightarrow Fe_2O_3\) dư
Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\\n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\end{matrix}\right.\)
`=>` \(m_X=\left(0,3-\dfrac{1}{15}\right).160+\dfrac{2}{15}.56=44,8\left(g\right)\)
b) \(n_{Fe_2O_3\left(dư\right)}=0,3-\dfrac{1}{15}=\dfrac{7}{30}\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Theo PT: \(n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=\dfrac{5}{3}\left(mol\right)\)
`=>` \(m_{ddHCl}=\dfrac{\dfrac{5}{3}.36,5}{3,65\%}=\dfrac{5000}{3}\left(g\right)\)
c) Theo PT: \(n_{H_2}=n_{Fe}=\dfrac{2}{15}\left(mol\right)\)
`=>` \(m_{ddspư}=\dfrac{5000}{3}+44,8-\dfrac{2}{15}.2=1711,2\left(g\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=\dfrac{2}{15}\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=\dfrac{7}{15}\left(mol\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{\dfrac{2}{15}.127}{1711,2}.100\%=1\%\\C\%_{FeCl_3}=\dfrac{\dfrac{7}{15}.162,5}{1711,2}.100\%=4,43\%\end{matrix}\right.\)
