Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
a. \(PTHH:Fe+H_2SO_4--->FeSO_4+H_2\uparrow\)
b. Theo PT: \(n_{H_2}=n_{Fe}=n_{H_2SO_4}=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,25.22,4=5,6\left(lít\right)\)
c. Ta có: \(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
\(\Rightarrow C_{\%_{H_2SO_4}}=\dfrac{24,5}{486,5}.100\%=5,04\%\)