When mole of $C o C l_{3} .6 N H_{3}$ is treated with excess of AgNO $_{3}$ solution. The number of moles of AgCl precipitated is:

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Solution

The correct option is B 3
One mole of $C o C l_{3} .6 N H_{3}$ gives 3 moles of $C l$ ion so when mole of $C o C l_{3} .6 N H_{3}$ is treated with excess of $A g N O_{3}$ solution, it gives 3 moles of $A g C l$ .

$C o C l_{3} .6 N H_{3} (a q) + e x c e s s A g N O_{3} (a q) \to 3 A g C l + C o (N O_{3})_{3} .6 N H_{3} (a q)$

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When mole of CoCl3.6NH3 is treated with excess of AgNO3 solution. The number of moles of AgCl precipitated is:

The correct option is B 3One mole of CoCl3.6NH3 gives 3 moles of Cl ion so when mole of CoCl3.6NH3 is treated with excess of AgNO3 solution, it gives 3 moles of AgCl.CoCl3.6NH3(aq)+excessAgNO3(aq)→3AgCl+Co(NO3)3.6NH3(aq)

When mole of $C o C l_{3} .6 N H_{3}$ is treated with excess of AgNO $_{3}$ solution. The number of moles of AgCl precipitated is: