# How to solve

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Correct option is C) Given, 1 KClO 3 ​ decomposes to give 1 mole O 2 ​ The reaction involved is 2KClO 3 ​ ⟶2KCl+3O 2 ​ It is clear from above equation that 2 moles KClO 3 ​ produce 3 moles O 2 ​ but it is given that 1 mole O 2 ​ is produced. ⇒ Moles of KClO 3 ​ used= 3 1×2 ​ =0.66 moles⟶(1) ∴ The left over moles of KClO 3 ​ are 1−0.66=0.34 moles ⟶(2) Now the parallel reaction is 4KClO 3 ​ ⟶3KClO 4 ​ +KCl It is clear that 4 moles KClO 3 ​ gives 3 moles KClO 4 ​ & 1 mole KCl⟶(3) Now, for determining mole fraction of KClO 4 ​ first calculate moles of KClO 4 ​ & KCl From (2) & (3) If 4 moles KClO 3 ​ gives 3 moles KClO 4 ​ ⇒0.34 moles KClO 3 ​ give 4 0.34×3 ​ ∴ No. of moles of KClO 4 ​ produced=0.255 moles Now, if 4 moles KClO 3 ​ give 1 mole KCl ⇒0.34 moles KClO 3 ​ give 4 0.34×1 ​ moles ∴ No. of Moles of KCl produced=0.085 moles ∴ Total no. of Moles in final mix=0.255+0.085 =0.34 moles ∴ Mole fraction of KClO 4 ​ = Totalmoles MolesofKClO 4 ​ ​ = 0.34 0.255 ​ =0.75
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