20 mL of 0.2 M Al_2(SO_4)_3 is mixed with | vedclass

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(B) The reaction between $Al_2(SO_4)_3$ and $BaCl_2$ is: $Al_2(SO_4)_3 + 3BaCl_2 \rightarrow 2AlCl_3 + 3BaSO_4(s)$.Initial moles of $BaCl_2 = 20 \ mL

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$6.6$

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(B) The reaction between $Al_2(SO_4)_3$ and $BaCl_2$ is: $Al_2(SO_4)_3 + 3BaCl_2 ightarrow 2AlCl_3 + 3BaSO_4(s)$.Initial moles of $BaCl_2 = 20 \ mL 	imes 6.6 \ M = 132 \ mmol$.Initial moles of $Al_2(SO_4)_3 = 20 \ mL 	imes 0.2 \ M = 4 \ mmol$.Since $BaCl_2$ is in excess,all $Al_2(SO_4)_3$ reacts with $12 \ mmol$ of $BaCl_2$ to form $8 \ mmol$ of $AlCl_3$ and $12 \ mmol$ of $BaSO_4$ precipitate.Remaining moles of $BaCl_2 = 132 \ mmol - 12 \ mmol = 120 \ mmol$.Total moles of $Cl^{-}$ ions = (moles from remaining $BaCl_2$) + (moles from $AlCl_3$ formed) = $(120 	imes 2) + (8 	imes 3) = 240 + 24 = 264 \ mmol$.Total volume of solution = $20 \ mL + 20 \ mL = 40 \ mL$.Concentration of $Cl^{-} = \frac{264 \ mmol}{40 \ mL} = 6.6 \ M$.

$20 \ mL$ of $0.2 \ M \ Al_2(SO_4)_3$ is mixed with $20 \ mL$ of $6.6 \ M \ BaCl_2$. The concentration of $Cl^{-}$ ion in the final solution is $......... \ M$.

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