PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a, Ta có: \(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)
Theo PT: \(n_{H_2}=3n_{Fe_2O_3}=0,225\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,225.22,4=5,04\left(l\right)\)
b, Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
Bạn tham khảo nhé!