Ta có nHCl = \(\dfrac{10,95}{36,5}\) = 0,3 ( mol )
CuO + 2HCl \(\rightarrow\) CuCl2 + H2O
x → 2x → x → x
Zno + 2HCl \(\rightarrow\) ZnCl2 + H2O
y → 2y → y → y
=> \(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
=> % CuO = \(\dfrac{64\times0.05}{12,1}\) . 100 \(\approx\) 26,5%
=> % ZnO = 100 - 26,5 = 73,5 %
=> MH2O = ( x + y ).18 = ( 0,05 + 0,1 ).18 = 2,7 (gam)