Question

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The quantity of potassium chlorate (molecular weight 122.5 ) required in grams to produce 33.6 litre of oxygen at normal temperature and pressure will be?
a) 122.5
b) 132.6
c) 245.0
d) 254.0​

Answer 1

Answer:

since , we don't know the whole product formed after reaction.

We will use Principal of Atom conservation.

Explanation:

KCL03 ====> O2

122.5. 33.6

Since O-atom is common, in both side.

POAC on O-atom,

3 X mole of KClO3 = 2 X mole of O2

3 X mass(KClO3)/122.5 == 2 X 33.6/22.4

[mole = volume /molar volume]

mass(KClO3) = 122.5 Gm required.

So, mark it as brainlist.

Answer 2

Since , we don't know the whole product formed after reaction.

We will use Principal of Atom conservation.

Explanation:

KCL03 ====> O2

122.5. 33.6

Since O-atom is common, in both side.

POAC on O-atom,

3 X mole of KClO3 = 2 X mole of O2

3 X mass(KClO3)/122.5 == 2 X 33.6/22.4

[mole = volume /molar volume]

mass(KClO3) = 122.5 Gm required.