Mixture X= 0.02 mole of [Co(NH3)5SO4]Br and 0.02 mole of [Co(NH3)5Br]SO4 was prepared in 2 L of solution. 1 L of mixture X + excess AgNO3 arrow Y 1 L of mixture X + excess BaCI2 arrow Z Number of moles of Y and Z are

Question

Mixture X= 0.02 mole of [Co(NH3)5SO4]Br and 0.02 mole of [Co(NH3)5Br]SO4 was prepared in 2 L of solution. 1 L of mixture X + excess AgNO3 arrow Y 1 L of mixture X + excess BaCI2 arrow Z Number of moles of Y and Z are (A) 0.01, 0.01 (B) 0.02, 0.01 (C) 0.01, 0.02 (D) 0.02, 0.02

Tardigrade Admin · Accepted Answer

In 1 L solution, there will be 0.01 mole of each [Co(NH3)5SO4] Br and [Co(NH3)5Br]SO4. Addition of excess of AgNO3 will give 0.01 mole of AgBr. Addition of excess of BaCl2 will give 0.01 mole of BaSO4.

Q. Mixture $X = 0.02$ mole of $[C o (N H_{3})_{5} S O_{4}] B r$ and $0.02$ mole of $[C o (N H_{3})_{5} B r] S O_{4}$ was prepared in $2 L$ of solution.
$1 L$ of mixture $X +$ excess $A g N O_{3} \to Y$
$1 L$ of mixture $X$ + excess $B a C I_{2} \to Z$
Number of moles of $Y$ and $Z$ are

0.01, 0.01

0.02, 0.01

0.01, 0.02

0.02, 0.02

In $1 L$ solution, there will be $0.01$ mole of each $[C o (N H_{3})_{5} S O_{4}] B r$ and $[C o (N H_{3})_{5} B r] S O_{4} .$ Addition of excess of $A g N O_{3}$ will give $0.01$ mole of $A g B r$ . Addition of excess of $B a C l_{2}$ will give $0.01$ mole of $B a S O_{4} .$

Q. Mixture X=0.02 mole of [Co(NH3​)5​SO4​]Br and 0.02 mole of [Co(NH3​)5​Br]SO4​ was prepared in 2L of solution. 1L of mixture X+ excess AgNO3​→Y 1L of mixture X + excess BaCI2​→Z Number of moles of Y and Z are