PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,1\cdot98=9,8\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)
1 1 1 1
0,1 0,1 0,1
a) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{H2SO4}=0,1.98=9,8\left(g\right)\)
b) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
Chúc bạn học tốt
