\(Fe2O3+3H2\rightarrow2Fe+3H2O\)
\(CuO+H2\rightarrow Cu+H2O\)
Ta có :
\(m_{Fe2O3}:m_{CuO}=3:7\)
\(m_{Fe2O3}=\frac{3.9,6}{10}=2,88\left(g\right)\)
\(nFe2O3=0,018\left(mol\right)\)
\(mCuO=9,6-2,88=6,72\left(g\right)\)
\(nCuO=0,084\left(mol\right)\)
\(mCu=0,084.64=5,376\left(g\right)\)
\(V_{H2}=\left(0,054+0,084\right).22,4=3,0912\left(l\right)\)
