When Cl2 is bubbled through aqueous KOH, a firework explosive (A) is formed along with KCl and H2O, Correct statement/s is/are about this reaction is

6KOH+3Cl2→KClO3(A)+5KCl+3H2O From , Pv=nRT P= 980760atm V= 150mL= 0.15L R=0.0891L.atm.mol-1K-1.T=298K n=PVRT=980 X 150 760 X 0.0821 X 298=7.9 mol moles of KClO3=13X moles of Cl2. = 13X7.9 = 2.633 moles Mass of KClO3 = 2.633 X 122.5 =322.58g

When Cl2 is bubbled through aqueous KOH, a firework explosive (A) is formed along with KCl and H2O, Correct statement/s is/are about this reaction is

6KOH+3Cl2→KClO3(A)+5KCl+3H2O From , Pv=nRT P= 980760atm V= 150mL= 0.15L R=0.0891L.atm.mol-1K-1.T=298K n=PVRT=980 X 150 760 X 0.0821 X 298=7.9 mol moles of KClO3=13X moles of Cl2. = 13X7.9 = 2.633 moles Mass of KClO3 = 2.633 X 122.5 =322.58g

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When Cl₂ is bubbled through aqueous KOH, a firework explosive (A) is formed along with KCl and H₂O, Correct statement/s is/are about this reaction is

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323g of A will be formed by 150L of Cl₂ at 980mm of Hg at 298K

7.9 moles of A is formed under the condition B option

Oxidation state of Cl in A is +5

Oxidation state of K in A is –1

answer is A, B.

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Detailed Solution

$6 KOH + 3 {Cl}_{2} \to \underset{(A)}{{KClO}_{3}} + 5 KCl + 3 H_{2} O$

$F r o m, P v = n R T P = \frac{980}{760} a t m V = 150 m L = 0.15 L R = 0.0891 L . a t m . m o l^{- 1} K^{- 1} . T = 298 K n = \frac{P V}{R T} = \frac{980 X 150}{760 X 0.0821 X 298} = 7.9 m o l m o l e s o f K C l O_{3} = \frac{1}{3} X m o l e s o f C l_{2} . = \frac{1}{3} X 7.9 = 2.633 m o l e s M a s s o f K C l O 3 = 2.633 X 122.5 = 322.58 g$