Đáp án:

$\% {m_{Zn}} = 8,28\% ;\% {m_{Na}} = 5,86\% ;\% {m_{N{a_2}O}} = 8\% ;\% {m_{BaO}} = 77,86\% $

$  {m_X} = 7,85g$

Giải thích các bước giải:

 Sơ đồ phản ứng: 

$X\left\{ \begin{gathered}
  Zn:a \hfill \\
  Na:b \hfill \\
  N{a_2}O:c \hfill \\
  BaO:d \hfill \\ 
\end{gathered}  \right.\xrightarrow{{ + {H_2}O}}\left\{ \begin{gathered}
  Y\left\{ \begin{gathered}
  N{a^ + } \hfill \\
  B{a^{2 + }} \hfill \\
  ZnO_2^{2 - } \hfill \\
  O{H^\_} \hfill \\ 
\end{gathered}  \right.\xrightarrow[{HCl:0,01}]{{{H_2}S{O_4}:0,06}}\left\{ \begin{gathered}
  9,815g\left\{ \begin{gathered}
  BaS{O_4} \hfill \\
  Zn{(OH)_2} \hfill \\ 
\end{gathered}  \right. \hfill \\
  3,52gZ\left\{ \begin{gathered}
  N{a^ + } \hfill \\
  SO_4^{2 - } \hfill \\
  C{l^ - } \hfill \\
  Z{n^{2 + }} \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered}  \right. \hfill \\
  {H_2}:0,02mol \hfill \\ 
\end{gathered}  \right.$

${n_{BaS{O_4}}} = d \Rightarrow {n_{SO_4^{2 - }(Z)}} = 0,06 - d$

Bảo toàn nguyên tố $Na$: ${n_{N{a^ + }(Z)}} = {n_{Na}} + 2{n_{N{a_2}O}} = b + 2c$

${n_{C{l^ - }}} = {n_{HCl}} = 0,1mol$

Bảo toàn điện tích dung dịch Z: $2{n_{SO_4^{2 - }(Z)}} + {n_{C{l^ - }}} = {n_{N{a^ + }}} + 2{n_{Z{n^{2 + }}}}$

$ \Rightarrow {n_{Z{n^{2 + }}}} = \dfrac{{2(0,06 - d) + 0,01 - (b + 2c)}}{2} = 0,065 - d - c - 0,5b$

$ \Rightarrow {m_Z} = {m_{N{a^ + }}} + {m_{C{l^ - }}} + {m_{SO_4^{2 - }}} + {m_{Z{n^{2 + }}}} = 3,52$

⇒ $23(b + 2c) + 0,355 + 96(0,06 - d) + 65.(0,065 - d - c - 0,5b) = 3,52$

$ \Rightarrow 9,5b + 19c + 161d = 6,82$ (1)

Bảo toàn nguyên tố $Zn$: ${n_{Zn}} = {n_{Zn{{(OH)}_2}}} + {n_{Z{n^{2 + }}(Z)}} \Rightarrow {n_{Zn{{(OH)}_2}}} = a + 0,5b+c+d-0,065$

$ \Rightarrow {m_ \downarrow } = {m_{BaS{O_4}}} + {m_{Zn{{(OH)}_2}}} = 9,815$

$ \Rightarrow 233d + 99(a + 0,5b + c + d - 0,065) = 9,815$

$ \Rightarrow 99a + 49,5b + 99c + 332d = 16,25$ (2)

$\% {m_O} = \dfrac{{16(c + d)}}{{65a + 23b + 62c + 153d}}.100\%  = 10,19\% $

$ \Rightarrow 6,6235a + 2,3437b - 9,6822c - 0,4093d = 0$ (3)

Bảo toàn electron: $2{n_{Zn}} + {n_{Na}} = 2{n_{{H_2}}} \Rightarrow 2a + b = 0,04$ (4)

Từ (1), (2), (3) và (4) $ \Rightarrow \left\{ \begin{gathered}
  a = 0,01 \hfill \\
  b = 0,02 \hfill \\
  c = 0,01 \hfill \\
  d = 0,04 \hfill \\ 
\end{gathered}  \right.$

$ \Rightarrow {m_X} = 0,01.65 + 0,02.23 + 0,01.62 + 0,04.153 = 7,85g$

$\begin{gathered}
   \Rightarrow \% {m_{Zn}} = \dfrac{{0,01.65}}{{7,85}}.100\%  = 8,28\%  \hfill \\
  \% {m_{Na}} = \dfrac{{0,02.23}}{{7,85}}.100\%  = 5,86\%  \hfill \\
  \% {m_{N{a_2}O}} = \dfrac{{0,01.62}}{{7,85}}.100\%  = 8\%  \hfill \\ 
\end{gathered} $

$\% {m_{BaO}} = 100 - 8,28 - 5,86 - 8 = 77,86\% $