Question

Answer 1

Gọi số mol CuO, Fe₂O₃ là a, b (mol)

=> 80a + 160b = 48 (1)

\(n_{H_2O}=\dfrac{13,5}{18}=0,75\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

               a---------------------->a

            Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

                 b------------------------>3b

=> a + 3b = 0,75 (2)

(1)(2) => a = 0,3 (mol); b = 0,15 (mol)

\(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,3.80}{48}.100\%=50\%\\\%m_{Fe_2O_3}=\dfrac{0,15.160}{48}.100\%=50\%\end{matrix}\right.\)