Answer 1

Ta có: \(n_{FeO}=\dfrac{7,2}{7,2}=0,1\left(mol\right)\)

\(m_{HCl}=300.7,3\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PT: \(FeO+2HCl\rightarrow FeCl_2+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{2}\), ta được HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{FeO}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{FeO}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,4\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 7,2 + 300 = 307,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127}{307,2}.100\%\approx4,1\%\text{ }\\C\%_{HCl\left(dư\right)}=\dfrac{0,4.36,5}{307,2}.100\%\approx4,75\%\end{matrix}\right.\)

Bạn tham khảo nhé!