\(CuO+H2-->Cu+H2O\)
\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Cu}=n_{H2}=0,15\left(mol\right)\)
\(m_{Cu}=0,15.64=9,6\left(g\right)\)
\(H\%=\frac{6,4}{9,6}.100\%=66,67\%\)
nH2= 0,15(mol); nCu= 0,1(mol)
PTHH: CuO + H2 -to-> Cu + H2O
0,15_________0,15____0,15(mol)
nCu(LT)=0,15(mol)
=> H= (0,1/0,15).100\(\approx66,667\%\)