nFe = \(\dfrac{33,6}{56}=0,6\left(mol\right)\)
Pt: Fe + H2SO4 --> FeSO4 + H2
.....0,6---> 0,6-------> 0,6---> 0,6.....(mol)
mFeSO4 = 0,6 . 152 = 91,2 (g)
VH2 = 0,6 . 22,4 = 13,44 (lít)
mdd H2SO4 dùng = \(\dfrac{0,6.98}{20}.100=294\left(g\right)\)
a) Fe + H2SO4 → FeSO4 + H2↑
b) \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
Theo pT: \(n_{FeSO_4}=n_{Fe}=0,6\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,6\times152=91,2\left(g\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,6\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,6\times22,4=13,44\left(l\right)\)
c) Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,6\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,6\times98=58,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{58,8}{20\%}=294\left(g\right)\)
PTHH. Fe + H2SO4 → FeSO4 + H2↑
Theo bài có: nFe=\(\dfrac{33,6}{56}\)=0,6(mol)
Theo pthh và bài ta có:
+)nFeSO4=nFe=0,6(mol)
⇒mFeSO4=0,6×152=91,2(g)
+) nH2=nFe=0,6(mol)
⇒VH2=0,6×22,4=13,44(l)
+) nH2SO4=nFe=0,6(mol)
⇒mH2SO4=0,6×98=58,8(g)⇒mddH2SO4=\(\dfrac{58,8}{20\%}\)=294(g)