Question

Làm cho em câu 3 trước với ạ 

Answer 1

Câu 3:

a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)

c, \(2Fe+6H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

Theo PT: \(n_{SO_2}=\dfrac{3}{2}n_{Fe}=0,225\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,225.22,4=5,04\left(l\right)\)