If 6 mol of metal is treated with 8 mol dil. HNO 3 followed by treating (B) with excess KCl then moles of white ppt. formed will be ________.

6 Hg + 8 HNO 3 → 3 Hg 2 ( NO 3 ) 2 + 2 NO + 4 H 2 O Hg 2 ( NO 3 ) 2 + 2 KCl → Hg 2 Cl 2 ↓ + 2 K + + 2 NO 3 −

Chemistry - Salt Analysis Question with Solution | TestHub

ChemistrySalt AnalysisTEST OF CATIONS Dry Test, Group Analysis, Group 1,2Medium2 minQB

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If 6 mol of metal is treated with 8 mol dil. HNO₃ followed by treating (B) with excess KCl then moles of white ppt. formed will be ________.

Answer:

Solution:

$6 Hg + 8 HNO_{3} \to 3 Hg_{2} (NO_{3})_{2} + 2 NO + 4 H_{2} O$ $Hg_{2} (NO_{3})_{2} + 2 KCl \to Hg_{2} Cl_{2} ↓ + 2 K^{+} + 2 NO_{3}^{-}$

Stream:JEESubject:ChemistryTopic:Salt AnalysisSubtopic:TEST OF CATIONS Dry Test, Group Analysis, Group 1,2

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