Volume of oxygen at STP which will be obtained by the action of heat on 20 gm of $K C l O_{3}$ is:
$(K = 39, C l = 35.5, O = 16)$

5.486 L

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11.2 L

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22.4 L

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2.74 L

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Solution

The correct option is A 5.486 L
Reaction:
$2 K C l O_{3} \to 2 K C l + 3 O_{2}$
2 2 3
So, volume of oxygen obtained $= 3 \times 22.4 \times \frac{20}{122.5} \times 2 = 5.486 L$

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Similar questions

20

K C l O_{3}

[K = 39, C l = 35.5, O = 16]

a) What is percentage composition of potassium chlorate (KClO₃).

Given:Atomic mass Of K=39,Cl=35.5andO=16

2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}

K C l O_{3}

O_{2}

2 K C l O_{3} Δ \to 2 K C l + 3 O_{2}; C + O_{2} Δ \to C O_{2}

K C l O_{3}

O_{2}

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