A mixture X of 0.02 , mol of [Co(NH_3)_5SO_4] | vedclass

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(C) The total volume of the mixture is $2 \, L$,containing $0.02 \, mol$ of each complex. Therefore,$1 \, L$ of the mixture contains $0.01 \, mol$ of

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$0.01, 0.01$

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(C) The total volume of the mixture is $2 \, L$,containing $0.02 \, mol$ of each complex. Therefore,$1 \, L$ of the mixture contains $0.01 \, mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.01 \, mol$ of $[Co(NH_3)_5Br]SO_4$.When $1 \, L$ of mixture $X$ reacts with excess $AgNO_3$,only the ionizable $Br^-$ reacts to form $AgBr$ precipitate:$[Co(NH_3)_5SO_4]Br + AgNO_3 	o [Co(NH_3)_5SO_4]NO_3 + AgBr(s)$Since there is $0.01 \, mol$ of $[Co(NH_3)_5SO_4]Br$,the moles of $AgBr$ $(Y)$ formed is $0.01 \, mol$.When $1 \, L$ of mixture $X$ reacts with excess $BaCl_2$,only the ionizable $SO_4^{2-}$ reacts to form $BaSO_4$ precipitate:$[Co(NH_3)_5Br]SO_4 + BaCl_2 	o [Co(NH_3)_5Br]Cl_2 + BaSO_4(s)$Since there is $0.01 \, mol$ of $[Co(NH_3)_5Br]SO_4$,the moles of $BaSO_4$ $(Z)$ formed is $0.01 \, mol$.Thus,the number of moles of $Y$ and $Z$ are $0.01$ and $0.01$ respectively.

$A$ mixture $X$ of $0.02 \, mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.02 \, mol$ of $[Co(NH_3)_5Br]SO_4$ was prepared in $2 \, L$ of solution:
$1 \, L$ of mixture $X +$ excess of $AgNO_3 \to Y$
$1 \, L$ of mixture $X +$ excess of $BaCl_2 \to Z$
Number of moles of $Y$ and $Z$ respectively are

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$A$ mixture $X$ of $0.02 \, mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.02 \, mol$ of $[Co(NH_3)_5Br]SO_4$ was prepared in $2 \, L$ of solution:$1 \, L$ of mixture $X +$ excess of $AgNO_3 \to Y$$1 \, L$ of mixture $X +$ excess of $BaCl_2 \to Z$Number of moles of $Y$ and $Z$ respectively are