a, Ta có nHCl = \(\dfrac{10,95}{36,5}\) = 0,3 ( mol )
CuO + 2HCl → CuCl2 + H2O
x → 2x → x → x
ZnO + 2HCl → ZnCl2 + H2O
y → 2y → y → y
=> \(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
=> mCuCl2 = 0,05 . 135 =6,75 ( gam )
=> mZnCl2 = 0,1 .136 = 13,6 ( gam )
b, mH2O = ( 0,05 + 0,1 ) . 18 = 2,7 ( gam )