\(R+H_2SO_4\rightarrow RSO_4+H_2\)
a) \(n_{H_2}=n_R=0,45\left(mol\right)\)
=> \(M_R=\dfrac{25,2}{0,45}=56\left(Fe\right)\)
b)\(n_{H_2SO_4}=n_{H_2}=0,45\left(mol\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{0,45.98}{9,8\%}=450\left(g\right)\)
c) \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{HCl}=2n_{Fe}=0,9\left(mol\right)\)
=> \(V_{HCl}=\dfrac{0,9}{2}=0,45\left(lít\right)=450ml\)
