Question

Find the mass of KCLO3 required to prepare 3.2 gm of oxygen

Answer 1

Answer:

The mass of KCLO3 required to prepare 3.2 gm of oxygen is 8.16g.

Explanation:

From the equation,

2 moles of KClO₃ gives 3 moles of O₂

2 KClO₃⇒ 3O₂

we know that,

atomic mass of potassium = 39g

atomic mass of Cl = 35.5

atomic mass of oxygen = 16

∴ molar mass of KClO₃ is 122.5g

2 moles of KClO₃ is 245g

so, 245g KClO₃ gives (3×32)= 96g O₂

that means, one gram of O₂ is prepared from = $\frac{245}{96}$ = 2.55g KClO₃

so, 3.2g O₂ is prepared from = 3.2×2.55

                                                  = 8.16g of KClO₃

Therefore, to prepare 3.2 g of oxygen we need 8.16 gram of KClO₃.

∴ the answer is 8.16gram.

Answer 2

Given: 3.2 grams of oxygen is prepared.

To find: Mass of KClO₃ required to prepare 3.2 gm of oxygen  

Solution:

The following reaction takes place when KClO₃ is dissociated -

2KClO₃ = KCl + KClO+O₂↑

Two moles of KClO₃ dissociated to produce one mole of oxygen.

One mole of oxygen is equal to 32 grams of oxygen.

Therefore, 3.2 grams of oxygen is equal to 0.1 moles of oxygen.

Therefore, 0.1 moles of oxygen is prepared from 0.2 moles of potassium chlorate.

One mole of Potassium chlorate is equal to 122 grams.

Therefore, 0.2 moles of Potassium chlorate will be equal to 122 × 0.2 = 24.4 grams of potassium chlorate.

So, the mass of KClO₃ required to prepare 3.2 grams of oxygen is 24.4 grams.