a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,05}{1}\) => Fe dư, H2SO4 hết
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,05<---0,05------>0,05--->0,05
=> V = 0,05.22,4 = 1,12 (l)
b)
PTHH: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
0,05->0,0375
=> mFe = 0,0375.56 = 2,1 (g)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,05}{1}\), ta được Fe dư.
Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,05.22,4=1,12\left(mol\right)\)
b, PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
____________0,05__0,0375 (mol)
\(\Rightarrow m_{Fe}=0,0375.56=2,1\left(g\right)\)
Bạn tham khảo nhé!
