a)
\(Fe+H_2SO_4-t^0\rightarrow FeSO_4+H_2\\ theogt:nFe=\dfrac{22,4}{56}=0,4\left(mol\right);n_{H2SO4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\ theoPTHH:n_{Fe}=1\left(mol\right);n_{H2SO4}=1\left(mol\right)\\ \)
b)ta có tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,25}{1}\Rightarrow Fe\) dư, tính số mol các chất theo H2SO4
theo PTHH\(n_{Fe}=n_{FeSO4}=n_{H2}=n_{H2SO4}=0,25\left(mol\right)\)
\(\Rightarrow n_{Fedư}=0,4-0,25=0,15\left(mol\right)\\ \Rightarrow m_{Fedư}=0,15\cdot56=8,4\left(g\right)\\ \)
c)\(n_{FeSO4}=0,25\left(mol\right)\Rightarrow m_{FeSO4}=0,25\cdot152=38\left(g\right)\)
d)\(n_{H2}=0,25\left(mol\right)\Rightarrow V_{H2}=0,25\cdot22,4=5,6\left(l\right)\)