PTHH: 

`Al->Al^(+3)+3e`

`Zn->Zn^(+2)+2e`

`O_2+4e->2O^(-2)`

`n_(HCl)=2n_(H_2)=(2*17,696)/(22,4)=1,58mol`

`n_(Cl^-)=n_(HCl)=1,58mol=>n_(O_2)=n_(Cl^-)/2=(1,58)/2=0,79mol`

`m_Y=m_X+m_(O_2^-)=17+16*0,79=29,64g`

Đáp án:

 29,64g

Giải thích các bước giải:

\(\begin{array}{l}
Al \to A{l^{ + 3}} + 3e\\
Zn \to Z{n^{ + 2}} + 2e\\
{O_2} + 4e \to 2{O^{ - 2}}\\
BT\,e:3{n_{Al}} + 2{n_{Zn}} = 4{n_{{O_2}}}(1)\\
Al \to A{l^{ + 3}} + 3e\\
Zn \to Z{n^{ + 2}} + 2e\\
2HCl + 2e \to {H_2}\\
{n_{{H_2}}} = \dfrac{{17,696}}{{22,4}} = 0,79\,mol\\
BT\,e:3{n_{Al}} + 2{n_{Zn}} = 2{n_{{H_2}}}\\
 \Rightarrow 3{n_{Al}} + 2{n_{Zn}} = 2 \times 0,79 \Rightarrow 3{n_{Al}} + 2{n_{Zn}} = 1,58(2)\\
\text{ Thay(2) vào (1)} \Rightarrow 4{n_{{O_2}}} = 1,58\,mol \Rightarrow {n_{{O_2}}} = \dfrac{{1,58}}{4} = 0,395\,mol\\
BTKL:m = {m_X} + {m_{{O_2}}} = 17 + 0,395 \times 32 = 29,64g
\end{array}\)