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$Na_2SO_3\overset{+H_2SO_4}{\rightarrow}SO_2$
$x x$
$NaHSO_3\overset{+H_2SO_4}{\rightarrow}SO_2$
$y y$
$SO_2+Br_2+2H_2O\rightarrow H_2SO_4+2HBr$
$NaHSO_3+NaOH\rightarrow Na_2SO_3+H_2O$
$X:\begin{cases}Na_2SO_3:x mol \\NaHSO_3:y mol\\Na_2SO_4:z mol \end{cases}\Rightarrow \begin{cases}126x+104y+142z=28,56 \\x+y=n_{Br_2}=0,135\\\frac{7,14y}{28,56}=n_{NaOH}=0,0027 \end{cases}\Rightarrow \begin{cases}x=0,1242 mol \\ y=0,0108 mol\\ \end{cases}$
$\Rightarrow \%Na_2SO_3=\frac{0,1242.126.100}{28,56}=54,79\%$
$\Rightarrow \%NaHSO_3=\frac{0,0108.104.100}{28,56}=3,93\%\%$
$\Rightarrow \%Na_2SO_4=(100-54,79-3,93)\%=41,28\%$
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Trả lời 16-03-13 12:34 AM
viet130480
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