a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{5,2}{65}=0,08\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,08\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,08.24,79=1,9832\left(l\right)\)
\(m_{ZnCl_2}=0,08.136=10,88\left(g\right)\)
c, \(n_{Fe_3O_4}=\dfrac{6,032}{232}=0,026\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,026}{1}>\dfrac{0,08}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,06\left(mol\right)\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
Đúng 1
Bình luận (0)
