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Mercuric iodate [Hg_(5)(IO_(6))_(2)] reacts witha mixture of KI and HCI according to the following equation . Hg_(5)(IO_(6))_(2) +34 KI +24 HCl to 5K_(2)HgI_(4) +8I_(2) +24KCl +12H_(2)O The liberated odine is treated with na_(2)S_(2)O_(3) solution , 1 mL of which isequivalentto 0.0499g of CuSO_(4).5H_(2)O.What volume( in mL ) of Na_(2)S_(2)O_(3) solutionwill be required to react with iodine liberated from 0.7245 g of [Hg_(5) (IO_(6))_(2)] ? [Hg = 200.5 , Cu = 63.5 ,I = 127 ] |
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Answer» Solution :From the stoichmetry of the given EQUATIONS , `Hg_(5) (IO_(6))_(2) +34 KI+24 HCI to 5K_(2)HgI_(4) +8I_(2) +24KCI +12H_(2)O` and `2Cuso_(4) .5H_(2)O + 4KI to Cu_(2)I_(2) + 2K_(2)SO_(4) +I_(2)` we have , ` 8 xx " moles of " Hg_(5)(IO_(6))_(2) = " moles of " I_(2)` and moles of `CuSO_(4) .5H_(2)O = 2 xx" moles of " I_(2)` ` :. " moles of " I_(1) = 1/2 xx " moles of " CuSO_(4) .5H_(2)O ` ` = 8 xx " moles of " Hg_(5) (IO_(6))_(2)` or moles of `CuSO_(4) .5H_(2)O = 16 xx` moles of `Hg (IO_(6))_(2)` `("wt . of " CuSO_(4).5H_(2)O)/(249.5) = 16 xx (0.7245)/(1448.5){{:(CuSO_(4).5H_(2)O=249.5),(Hg_(5)(IO_(6))_(2)=1448.5):}}` ` :. ` wt of `CuSO_(4) .5H_(2)O = 1.9967 G ` Since `0.0499` g of `CuSO_(4) . 5H_(2) = 1 ` mL of `Na_(2)S_(2)O_(3)` ` :.1.9967 " g of " CuSO_(4) . 5H_(2)O = (1.9967)/(0.0499) = 40` mL |
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