a)\(Fe+H2SO4-->FeSO4+H2\)
\(n_{H2SO4}=\frac{12,25}{98}=0,125\left(mol\right)\)
\(n_{Fe}=n_{H2SO4}=0,125\left(mol\right)\)
\(m_{Fe}=0,125.56=7\left(g\right)\)
b)\(n_{H2}=n_{H2SO4}=0,125\left(mol\right)\)
\(V_{H2}=0,125.22,4=2,8\left(l\right)\)
nH2SO4=0,125(mol)
Fe + H2SO4 -> FeSO4 + H2
0,125<= 0,125 =>0,125
mFe=56.0,125=7(g)
VH2=22,4.0,125=2,8(lít)