Ta có: \(n_{H_2}=\dfrac{1,22}{22,4}\approx0,05\left(mol\right)\)
a. PTHH: Fe + H2SO4 ---> FeSO4 + H2
b. Theo PT: \(n_{Fe}=n_{H_2}=0,05\left(mol\right)\)
=> \(m_{Fe}=0,05.56=2,8\left(g\right)\)
c. Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,05\left(mol\right)\)
=> \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)