\(Fe2O3+3H2-->2Fe+3H2O\)
\(nH2=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{FE2O3}=\frac{32}{160}=0,2\left(mol\right)\)
\(\frac{0,3}{3}< \frac{0,2}{1}\Rightarrow Fe2O3dư\)
\(n_{Fe}=\frac{2}{3}n_{H2}=0,2\left(mol\right)\)
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(Fe_2O_3+3H_2O\rightarrow^{t^0}2Fe+3H_2O\)
1:3 (tỉ lệ)
\(n_{Fe_2O_3}=\frac{32}{160}=0.2\left(mol\right)\)
\(n_{H_2}=\frac{6.72}{22.4}=0.3\left(mol\right)\)
So sánh: \(\frac{0.2}{1}>\frac{0.3}{3}\)
=> \(Fe_2O_3\) dư, bài toán tính theo \(H_2\)
=> \(n_{Fe}=\frac{2}{3}\cdot n_{H_2}=\frac{2}{3}\cdot0.3=0.2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0.2\cdot56=11.2\left(g\right)\)
