a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,5.24,79=12,395\left(l\right)\)
\(m_{ZnCl_2}=0,5.136=68\left(g\right)\)
c, \(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,5}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,375\left(mol\right)\Rightarrow m_{Fe}=0,375.56=21\left(g\right)\)
Đúng 1
Bình luận (0)
