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Question

hon hop A gom S va Mg.dun nong A trong dieu kien khong co khong khi,sau do lam nguoi va cho san pham tac dung voi dung dich HCl (du) thu duoc 2,987(l) khi Bco ti khoi so voi khong khi bang 0,8966.dot chay het khi B sau do cho toan bo san pham vao 100 ml H2O2 5%(D=1) thu duoc dung dich Dxac dinh % khoi luong cac chat trong A va nong do % cac chat trong dung dich D

2,987 lit ở đk nào vậy ? Sao cho số liệu lẻ vậy ?

viet130480 · Answer 1 · 6/13/2013 2:32:02 PM

$Mg+S\rightarrow MgS$

$y y y$

$Mg+2HCl\rightarrow MgCl_2+H_2\uparrow$

$x x$

$MgS+2HCl\rightarrow MgCl_2+H_2S\uparrow$

$y y$

$\begin{cases}H_2:x mol \\H_2S:y mol \end{cases}\Rightarrow \begin{cases}x+y=\frac{2,987}{22,4} \\ \frac{2x+34y}{x+y}=29.0,8966 \end{cases}\Rightarrow \begin{cases}x=0,03335 mol \\ y=0,1 mol \end{cases}$

$\Rightarrow A:\begin{cases}\%S=\frac{32y}{32y+24(x+y)}.100\approx50\% \\\%Mg\approx 50\%\end{cases}$

$*)n_{H_2O_2}=0,147... mol$

$2H_2+O_2\rightarrow 2H_2O$

$x x$

$2H_2S+3O_2\rightarrow 2H_2O+2SO_2$

$y y y$

$SO_2+H_2O_2\rightarrow H_2SO_4$

$\Rightarrow n_{H_2SO_4}=n_{SO_2}=0,1 mol\Rightarrow m_{H_2SO_4}=9,8 gam$

$\Rightarrow m_{H_2O_2 du}=5-0,1.34=1,6 gam$

$*)m_D=1.100+18(x+y)+64y=108,8 gam$

$\Rightarrow \begin{cases}\%m_{H_2SO_4}=9\% \\ \%m_{H_2O_2}=1,47\% \end{cases}$