Đáp án đúng:
Giải chi tiết:1.
\(x\left\{ \begin{gathered} Mg \hfill \\ Zn \hfill \\ \end{gathered} \right. + \left\{ \begin{gathered} HCl:0,2 \hfill \\ {H_2}S{O_4}:0,04 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} {H_2} \uparrow \hfill \\ dd\,Y \to \left[ \begin{gathered} \xrightarrow[{0,3}]{{ + NaOH}} \downarrow :8,43\,g \hfill \\ \left\{ \begin{gathered} KOH:0,4V \hfill \\ Ba{(OH)_2}:0,05V \hfill \\ \end{gathered} \right.\xrightarrow{{}} \downarrow \xrightarrow{{{t^0},kk}}Ran:m(g) \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.\)
Giả sử ban đầu mol Mg: x và Zn: y
\(ddY\xrightarrow[{0,3}]{{ + NaOH}}ddA\left\{ \begin{gathered} \xrightarrow{{BTNT\,:\,\,Cl}}NaCl:0,2 \hfill \\ \xrightarrow{{BTNT:\,S{O_{4\,}}}}N{a_2}S{O_4}:0,04 \hfill \\ \xrightarrow{{BTNT:\,Na}}N{a_2}Zn{O_2}:\frac{{0,3 - 0,2 - 2.0,04}}{2} = 0,01(mol) \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} Mg{(OH)_2}:x \hfill \\ Zn{(OH)_2}:y - 0,01 \hfill \\ \end{gathered} \right.\)
\(\begin{gathered} = > hpt\left\{ \begin{gathered} 24x + 65y = 5,34 \hfill \\ 58x + 99(y - 0,01) = 8,43 \hfill \\ \end{gathered} \right. = > \left\{ \begin{gathered} x = 0,06 \hfill \\ y = 0,06 \hfill \\ \end{gathered} \right. \hfill \\ dd\,Y + \left\{ \begin{gathered} HCl:0,4V \hfill \\ {H_2}S{O_4}:0,05V \hfill \\ \end{gathered} \right. \to dd\left\{ \begin{gathered} \xrightarrow{{BTNT:Cl}}KCl:0,2 \hfill \\ \xrightarrow{{BTNT:\,S{O_4}}}{K_2}S{O_4}:0,04 - 0,05V \hfill \\ \end{gathered} \right. \hfill \\ \xrightarrow{{BTNT:K}}\left\{ \begin{gathered} nKOH = nKCl + 2n{K_2}S{O_4} \hfill \\ \to 0,4v = 0,2 + 0,04 - 0,05V \hfill \\ \end{gathered} \right. \to V = 0,56 \to \downarrow \left\{ \begin{gathered} Mg{(OH)_2}:0,06 \hfill \\ Zn{(OH)_2}:0,06 \hfill \\ BaS{O_4}:0,028 \hfill \\ \end{gathered} \right. \to Ran\left. {\left\{ \begin{gathered} MgO:0,06 \hfill \\ ZnO:0,06 \hfill \\ BaS{O_4}:0,028 \hfill \\ \end{gathered} \right.} \right\} \to m = 13,784\,g \hfill \\ \end{gathered} \)
Vậy giá trị của m = 13,784 g
Gọi CM của dung dịch H2SO4 là x (M) và B1: y1 (M); B2: y2 (M)
2NaOH + H2SO4 → Na2SO4 + 2H2O
Ta có
\(\left\{ \begin{gathered} \left. \begin{gathered} 0,01{y_1} + 0,01{y_2} = 0,04x \hfill \\ 0,02{y_1} + 0,01{y_2} = 0,065x \hfill \\ \end{gathered} \right\} \to \left\{ \begin{gathered} {y_1} = 2,5x \hfill \\ {y_2} = 1,5x \hfill \\ \end{gathered} \right. \hfill \\ \frac{{0,07a}}{{a + b}}{y_1} + \frac{{0,07b}}{{a + b}}{y_2} = 0,135x \hfill \\ \end{gathered} \right.\xrightarrow{{}}\frac{a}{b} = \frac{3}{4}\)
Vậy tỉ lệ a: b = 3: 4
