Ta có:
\(m_{Fe_2O_3}\) = mhh. 80% = 25 . 80% = 20(g) => \(n_{Fe_2O_3}\) = \(\dfrac{m}{M}=\dfrac{20}{160}=0,125\left(mol\right)\)
=>\(m_{CuO}\) = mhh - mFe2O3 = 25 - 20 = 5(g) => \(n_{CuO}\) = \(\dfrac{m}{M}=\dfrac{5}{80}=0,0625\left(mol\right)\)
Ta có: \(\Sigma n_O=3nFe_2O_3+nCuO=3.0,125+0,0625=0,4375\left(mol\right)\)
PT:
\(H_2+O\rightarrow H_2O\)
\(\Rightarrow n_O=n_{H_2}=0,4375\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,4375.22,4=9,8\left(l\right)\) \(\Rightarrow B\)