A mixture X containing 0.02 mol of [Co(NH_3) | vedclass

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(A) The mixture $X$ has $0.02 \ mol$ of each complex in $2 \ L$,so $1 \ L$ contains $0.01 \ mol$ of each complex.Reaction with $AgNO_3$: Only $[Co(NH_

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$0.01, 0.01$

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(A) The mixture $X$ has $0.02 \ mol$ of each complex in $2 \ L$,so $1 \ L$ contains $0.01 \ mol$ of each complex.Reaction with $AgNO_3$: Only $[Co(NH_3)_5SO_4]Br$ reacts with $AgNO_3$ to form $AgBr$ precipitate.$[Co(NH_3)_5SO_4]Br + AgNO_3 	o [Co(NH_3)_5SO_4]NO_3 + AgBr(s)$Since $0.01 \ mol$ of $[Co(NH_3)_5SO_4]Br$ is present in $1 \ L$,moles of $Y (AgBr) = 0.01 \ mol$.Reaction with $BaCl_2$: Only $[Co(NH_3)_5Br]SO_4$ reacts with $BaCl_2$ to form $BaSO_4$ precipitate.$[Co(NH_3)_5Br]SO_4 + BaCl_2 	o [Co(NH_3)_5Br]Cl_2 + BaSO_4(s)$Since $0.01 \ mol$ of $[Co(NH_3)_5Br]SO_4$ is present in $1 \ L$,moles of $Z (BaSO_4) = 0.01 \ mol$.Therefore,the number of moles of $Y$ and $Z$ are $0.01$ and $0.01$ respectively.

$A$ mixture $X$ containing $0.02 \ mol$ 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 $AgNO_3 \to Y$.
$1 \ L$ of mixture $X +$ excess $BaCl_2 \to Z$.
The number of moles of $Y$ and $Z$ are:

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$A$ mixture $X$ containing $0.02 \ mol$ 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 $AgNO_3 \to Y$.$1 \ L$ of mixture $X +$ excess $BaCl_2 \to Z$.The number of moles of $Y$ and $Z$ are: