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When an orange coloured crystalline compound (A) was heated with common salt and concentrated sulphuric acid an orange-yellow coloured gas (B) was evolved. The gas (B) when passed through caustic soda solution gave a yellow solution (C) which in turn gave the following reactions. (i) Addition of silver nitrate solution to (C) gave first a white precipitate which then turns red. Quantitatively, 0.155 g of the gas (B) required 2.0 m moles of AgNO_(3) to produce the first trace of red colour. (ii) Acidification of the solution (C) with dil. H_(2)SO_(4) gave an orange solution which contained chromium in +6 oxidation state. The solution liberated iodine from aqueous potassium iodide, leaving a green solution containing chromium in +3. oxidation state. Quantitatively, 0.155 g of the gas (B) liberated 1.5 m moles of iodine. Deduce the formula of A, B and C and explain the reactions. |
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Answer» Solution :Let us summarize the given reactions. The above set of reactions INDICATES that compound A is `K_(2)Cr_(2)O_(7)`, B is chromyl chloride gas and C is sodium chromate which explains all the given reactions as below. `underset(A("Orange crystals"))(K_(2)Cr_(2)O_(7))+4NaCl+6H_(2)SO_(4)to4NaHSO_(4)+2KHSO_(4)+underset("B(Orange)")(2CrO_(2)Cl_(2))uarr+3H_(2)O` `CrO_(2)Cl_(2)+4NaOHtounderset(C(Yellow))(Na_(2)CrO_(4))+2NaCl+2H_(2)O` `Na_(2)CrO_(4)+2AgNO_(3)tounderset(Red)(Ag_(2)CrO_(4))darr+2NaNO_(3)` `2Na_(2)CrO_(4)+H_(2)SO_(4)toNa_(2)Cr_(2)O_(7)+Na_(2)SO_(4)+H_(2)O` `Cr_(2)O_(7)^(2-)+14H^(+)+6l^(-)to2Cr^(3+)+7H_(2)O+3l_(2)` The quantitative data is explained in the following manner. From the above reactions we observe that `underset(155g)(CrO_(2)Cl_(2))=Na_(2)CrO_(4)=underset("2 moles")(2AgNO_(3))` 155g of `CrO_(2)Cl_(2)` require 2 moles of `AgNO_(3)` 0.155g of `CrO_(2)Cl_(2)` will require =`(2)/(155)xx0.1555=0.002 "moles" = 2m "moles"` This coincides with the given data. Similarly,`underset(2xx155)(2CrO_(2)Cl_(2))-=2CrO_(4^(2-))-=Cr_(2)O_(7)^(2-)-=underset("3 moles")(3l_(2)` Thus,`2xx155g` of `CrO_(2)Cl_(2)` liberate 3 moles of `l_(2)` 0.155g `CrO_(2)Cl_(2)` will liberate =`(3xx0.155)/(2xx155)=0.0015 "moles" =1.5m "moles"` This also coincides with the given data. 
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