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An aqueous solution $FeSO_4.Al_2(SO_4)_3$ and chrome alum is heated with excess of $Na_2O_2$ and filtered. The materials obtained are

(a) A colourless filtrate and a green residue.
(b) A yellow filtrate and a green residue.
(c) A yellow filtrate and a brown residue.
(d) A green filtrate and a brown residue.
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The correct option is (C) because
$Na_2O_2 + H_2O \longrightarrow NaOH + H_2O_2$
We have ions $Fe^{2+}, Al^{3+}$ and $Cr^{3+}$. Out of them $Fe^{2+}$ will be oxidized by $H_2O_2$ into $Fe^{3+}$ and $Cr^{3+}$ will be oxidized by $H_2O_2$ into $CrO_4^{2–}$ (medium being basic)
Then in presence of NaOH $Fe^{3+}$ will form $Fe(OH)_3$, insoluble in NaOH (Red brown ppt.).
But $Al^{3+}$ will form first $Al(OH)_3$ and then soluble $[Al(OH)_4]^–$ colorless.
$CrO_4^{2–}$ are soluble giving yellow solution. Hence, filterate will appear yellow and residue red brown.
answered Jan 26, 2014 by mosymeow_1
edited Apr 3, 2014 by mosymeow_1
 

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