Question

24g of a mixture of KCLO3 and KCL on being heated produces 7.6 g of oxygen. calculate the percentage of KCLO3 in the mixture.​

Answer 1

Explanation:

i hope this ans is right.

Answer 2

Answer:

Percentage of KClO3 in the mixture = 80.83 %

Explanation:

Given:

• Mass (Amount) of KClO3 and KCl mixture = 24 g
• Mass (Amount) of Oxygen produced = 7.6 g

To find:

• Percentage of KClO3 in the mixture = ?

Proof:

According to the question, chemical equation of the reaction:

KClO3 + KCl -------------Δ-------------> 7.6 g O2          

[Δ = Heat, KClO3 + KCl = 24g]

Note: As per the question, we want Oxygen, O2 which can 'only be obtained from KClO3'

∴ The balanced chemical reaction so as to obtain the result

= 2 KClO3 -------------Δ------------->2 KCl + 3O2

By the equation,

⇒ 2 (122.5)g KClO3 = 3 (32)g O2

[Molecular Mass of KClO3 = 39+ 35.5 + 16*3 = 122.5 & O2 = 16* 2= 32]

⇒     ------   x gram = 7.6 g O2

⇒ x = $\frac{7.6 X 2 X 122.5 }{3 X 32 }$

⇒ x = $= 19.4 g (Approx.)$

Thus, 7.6 g of O2 is produced from 19.4 g of KClO3

Hence,

Percentage of KClO3 in the mixture

= $\frac{Mass of KClO3}{Mass of the mixture} X 100$

= $\frac{19.4}{24} X 100$

= $\frac{9.7}{12} X 100$

= $\frac{9.7}{3} X 25$

= $\frac{242.5}{3}$

= 80.83 %

Hope you got that

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