Bài 2: Ta có: \(m_{Fe_2O_3}=\dfrac{m_{hh}.75}{100}=\dfrac{16.75}{100}=12\left(g\right)\)
=> mCuO=16-12=4(g)
=> nCuO=m/M=4/80=0,05(mol)
nfe2o3=m/M=0,075(mol)
PT1: CuO + H2 -t0-> Cu + H2O
vậy: 0,05--------------->0,05(mol)
=> mCu=n.M=0,05.64=3,2(g)
PT2: Fe2O3 + 3H2->2 Fe + 3H2O
vậy: 0,075------------>0,15(mol)
=> mfe=n.M=0,15.56=8,4(g