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Assigning Oxidation Numbers

67.2 L of O2 ...

Question

67.2 L of $O_{2}$ gas is collected at STP. If 4 mol of $K C l O_{3}$ is used in the following reactions:
$K C l O_{3} (s) ⟶ K C l (s) + O_{2} (g)$
$K C l O_{3} ⟶ K C l O_{4} + K C l$
The number of moles of $K C l O_{4}$ formed would be:

1.5 m o l

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1.2 m o l

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0.25 m o l

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3 m o l

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Solution

The correct option is A $1.5 m o l$
Moles of Oxygen gas produced
$= \frac{Volume(L) at STP}{22.4 L} = \frac{67.2}{22.4} = 3 m o l$

Applying POAC for oxygen atom,
$M o l e s o f O a t o m i n O_{2} = M o l e s o f O a t o m i n K C l O_{3}$
$2 \times n_{O_{2}} = 3 \times n_{K C l O_{3}}$
$n_{K C l O_{3}} = \frac{6}{3} m o l$
So $4 - 2 = 2$ moles of $K C l O_{3}$ remain for the second reaction.
For second reaction, applying POAC for oxygen atom,
$3 \times n_{K C l O_{3}} = 4 \times n_{K C l O_{4}}$
$3 \times 2 = 4 \times n_{K C l O_{4}}$
$n_{K C l O_{4}} = 1.5 m o l$

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Similar questions

O_{2}

K C l O_{3}

K C l O_{3} (s) ⟶ K C l (s) + O_{2} (g)

K C l O_{3} ⟶ K C l O_{4} + K C l

K C l O_{4}

O_{2}

K C l O_{3}

K C l O_{3} (s) ⟶ K C l (s) + O_{2} (g)

K C l O_{3} ⟶ K C l O_{4} + K C l

K C l O_{4}

K C l O_{3}

O_{2}

K C l O_{4}

(Molar mass of K C l O_{3} = 122.5 g m o l^{- 1}, K C l O_{4} = 138.5 g m o l^{- 1}) K C l O_{3} \to K C l + O_{2} K C l O_{3} \to K C l O_{4} + K C l

K C l O_{3}

K C l O_{3}

O_{2}

K C l O_{4}

K C l

K C l O_{4}

K O H

K C l O_{4}

K C l O_{4} = 138.5

g m o l^{- 1}

K O H = 56

g m o l^{- 1}

C l_{2} + 2 K O H \to K C l + K C l O + H_{2} O

3 K C l O \to 2 K C l + K C l O_{3}

4 K C l O_{3} \to K C l + 3 K C l O_{4}

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