Question

Answer 1

a) \(m_{CuO}=\dfrac{25.16}{100}=4\left(g\right)\) => \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{16-4}{160}=0,075\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

           0,05-->0,05--->0,05

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

           0,075-->0,225----->0,15

=> \(\left\{{}\begin{matrix}m_{Fe}=0,15.56=8,4\left(g\right)\\m_{Cu}=0,05.64=3,2\left(g\right)\end{matrix}\right.\)

b) \(n_{H_2}=0,05+0,225=0,275\left(mol\right)\)

=> \(V_{H_2}=0,275.22,4=6,16\left(l\right)\)