How many moles of AgCl would be obtained,when | vedclass

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(B) The coordination complex is $[Co(NH_3)_5Cl]Cl_2$. In this complex,there are $2$ ionizable $Cl^-$ ions outside the coordination sphere.The reaction

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$0.02$

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(B) The coordination complex is $[Co(NH_3)_5Cl]Cl_2$. In this complex,there are $2$ ionizable $Cl^-$ ions outside the coordination sphere.The reaction with excess $AgNO_3$ is:$[Co(NH_3)_5Cl]Cl_2 + 2AgNO_3 ightarrow [Co(NH_3)_5Cl](NO_3)_2 + 2AgCl \downarrow$Number of moles of complex $= M 	imes V(L) = 0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$.Since $1 \ mole$ of the complex yields $2 \ moles$ of $AgCl$,then $0.01 \ mole$ of the complex will yield $0.01 	imes 2 = 0.02 \ moles$ of $AgCl$.

How many moles of $AgCl$ would be obtained,when $100 \ mL$ of $0.1 \ M$ $[Co(NH_3)_5Cl]Cl_2$ is treated with excess of $AgNO_3$?

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An aqueous solution of $1 \ mol$ of the coordination compound $[Co(NH_3)_5Cl_3]$ gives $3 \ mol$ of ions. If $1 \ mol$ of the same compound reacts with $2 \ mol$ of $AgNO_3$ to produce $2 \ mol$ of $AgCl_{(s)}$,what is the correct formula of the coordination compound?

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