a, PT: \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
Ta có: \(n_{MnO_2}=\dfrac{30,45}{87}=0,35\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Cl_2}=n_{MnO_2}=0,35\left(mol\right)\\n_{HCl}=4n_{MnO_2}=1,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Cl_2}=0,35.71=24,85\left(g\right)\)
\(V_{ddHCl}=\dfrac{1,4}{1}=1,4\left(l\right)\)
b, PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
Ta có: \(n_{Fe}=\dfrac{25,2}{56}=0,45\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,45}{2}>\dfrac{0,35}{3}\), ta được Fe dư.
Theo PT: \(n_{FeCl_3}=\dfrac{2}{3}n_{Cl_2}=\dfrac{7}{30}\left(mol\right)\)
\(\Rightarrow m_{FeCl_3}=\dfrac{7}{30}.162,5\approx37,9\left(g\right)\)
Bạn tham khảo nhé!