Ta có: \(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{0,32}{56.2+16.3}=0,002\left(mol\right)\)
\(Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)
\(0,002\rightarrow0,006\rightarrow0,004\rightarrow0,006\) (mol)
a) \(m_{Fe}=n.M=0,004.56=0,224\left(g\right)\)
b) \(V_{CO_2}=n.22,4=0,006.22,4=0,1344\left(l\right)\)
\(a)Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)
\(1mol\) \(2mol\) \(3mol\)
\(0,002mol\) \(0,004mol\) \(0,006mol\)
\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{0,32}{160}=0,002\left(mol\right)\)
\(m_{Fe}=n.M=0,004.56=0,224\left(g\right)\)
\(\text{b)}V_{CO_2}=n,22,4=0,006.22,4=0,1344\left(l\right)\)