Question

Answer 1

a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,2<------------------0,2

\(\%m_{Fe}=\dfrac{0,2.56}{19,2}.100\%=58,33\%\)

\(\%m_{CuO}=\dfrac{19,2-0,2.56}{19,2}.100\%=41,67\%\)

b)

\(n_{CuO}=\dfrac{19,2-0,2.56}{80}=0,1\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,1-->0,1

=> V_H2 = 0,1.22,4 = 2,24 (l)