\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 → FeSO4 + H2
a) Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)
b) \(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2SO_4}\)
Theo bài: \(n_{Fe}=\frac{1}{2}n_{H_2SO_4}\)
Vì \(\frac{1}{2}< 1\) ⇒ H2SO4 dư, tính theo Fe
Theo pT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)
a) PTHH: Fe + H2SO4 → FeSO4 + H2
nFe=5,656=0,1(mol)nFe=5,656=0,1(mol)
Theo PT: nH2=nFe=0,1(mol)nH2=nFe=0,1(mol)⇒VH2=0,1×22,4=2,24(l)⇒VH2=0,1×22,4=2,24(l)
b) Theo PTHH:
nH2SO4=0,2×1=0,2(mol)
Theo PT: nFe=nH2SO4
Theo bài: nFe=12nH2SO4
Vì 12<112<1 ⇒ H2SO4 dư, tính theo Fe
Theo pT: nH2=nFe=0,1(mol)
⇒VH2=0,1×22,4=2,24(l)
