Bảo toàn electron

N

nguyenvancuong1225@gmail.com


xmol S, ymol $Br_2$

Ta có : $m_{S} + m_{Br_2}$ = 53,15 - ( 9,75+ 6,4 + 9) = 28 gam

$Zn \rightarrow Zn^{+2} + 2e$
0,15-----------------------------0,3mol
$Cu \rightarrow Cu^{+2} + 2e$
0,1--------------------------------0,2mol
$Ca \rightarrow Ca^{+2} + 2e$
0,225-----------------------------0,45mol
$\rightarrow n_{e cho} = 0,95 mol$

$S + 2e \rightarrow S^{-2}$
x----2xmol
$Br_2 + 2e \rightarrow 2Br^{-1}$
y---------2ymol
$\rightarrow n_{e nhận} = (2x + 2y) mol$

ta có
$\left\{\begin{matrix} 32x +160y = 28 \\ 2x + 2y = 0.95 \end{matrix}\right.$

\Leftrightarrow $ \left\{\begin{matrix} x = 0,375 mol \\ y = 0,1 mol \end{matrix}\right.$

$\rightarrow$ $m_S = 0,375.32 = 12 gam$

=)) =))
 
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