PTHH: Cu + 1/2 O2 -to-> CuO
x_________0,5x___x(mol)
Zn + 1/2 O2 -to-> ZnO
y________0,5y___y(mol)
Ta có: \(\left\{{}\begin{matrix}64x+65y=22,6\\80x+81y=28,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)
mCu=0,15.64=9,6(g)
=> %mCu= \(\frac{9,6}{22,6}.100\approx42,478\%\\ =>\%mZn\approx100\%-42,478\%\approx57,522\%\)
b) mO2= m(oxit)- m(kl)= 28,2- 22,6= 5,6(g)
=> nO2= 5,6/32=0,175(mol)
=>V(O2,đktc)= 0,175.22,4= 3,92(l)