a, \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,25\left(mol\right)\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)
b, \(H_2+O_{\left(trongoxit\right)}\rightarrow H_2O\)
\(n_{O\left(trongoxit\right)}=n_{H_2}=0,25\left(mol\right)\)
Có: mX giảm = mO (trong oxit) = 0,25.16 = 4 (g) = a