a. \(n_{H_2SO_4}=\dfrac{m_{H_2SO_4}}{M_{H_2SO_4}}=\dfrac{24,5}{98}=0,25\left(mol\right)\)
\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
- Mol theo PTHH : \(1:1:1:1\)
- Mol theo phản ứng : \(0,25\leftarrow0,25\rightarrow0,25\rightarrow0,25\)
\(\Rightarrow n_{Fe}=0,25\left(mol\right)\)
\(\Rightarrow m_{Fe}=n_{Fe}.M_{Fe}=0,25.56=14\left(g\right)\)
b. Từ a. suy ra \(n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,25.22,4=5,6\left(l\right)\)
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