Fe3O4 + 4CO \(\underrightarrow{to}\) 3Fe + 4CO2
\(n_{CO_2}=\frac{17,6}{44}=0,4\left(mol\right)\)
Theo pT: \(n_{Fe}=\frac{3}{4}n_{CO_2}=\frac{3}{4}\times0,4=0,3\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,3\times56=16,8\left(g\right)< 24,7\left(g\right)\)
Chất rắn sau phản ứng: Fe và Fe3O4 dư
\(\Rightarrow m_{Fe_3O_4}dư=24,7-16,8=7,9\left(g\right)\)
Theo PT: \(n_{Fe_3O_4}pư=\frac{1}{4}n_{CO_2}=\frac{1}{4}\times0,4=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}pư=0,1\times232=23,2\left(g\right)\)
Vậy \(m=m_{Fe_3O_4}dư+m_{Fe_3O_4}pư=7,9+23,2=31,1\left(g\right)\)
\(Fe_3O_4+4CO\rightarrow3Fe+4CO_2\uparrow\)
Chất rắn tạo thành có thể còn có Fe3O4
\(n_{CO_2}=\frac{17,6}{44}=0,4\left(mol\right)\)
\(TheoPT:n_{Fe}=\frac{3}{4}n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
\(\Rightarrow m_{Fe_3O_4\left(dư\right)}=24,7-16,8=7,9\left(g\right)\)
\(TheoPT:n_{Fe_3O_4}=\frac{1}{4}n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4\left(p.ứ\right)}=0,1.232=23,2\left(g\right)\)
\(\Rightarrow\Sigma m_{Fe_3O_4}=23,2+7,9=31,1\left(g\right)\)
