Question

How many grams of 02 will be formed from 6.5 moles of KCLO3

Answer 1

Answer:

312 g of O₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 mole of KClO₃ decomposed to 3 moles of O₂.

Next, we shall determine the number of mole of O₂ produced by the reaction of 6.5 moles of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 mole of KClO₃ decomposed to 3 moles of O₂.

Therefore, 6.5 moles of KClO₃ will decompose to produce = (6.5 × 3)/2 = 9.75 moles of O₂.

Finally, we shall determine the mass of 9.75 moles of O₂. This can be obtained as follow:

Mole of O₂ = 9.75 moles

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ =?

Mole = mass / Molar mass

9.75 = Mass of O₂ / 32

Cross multiply

Mass of O₂ = 9.75 × 32

Mass of O₂ = 312 g

Thus, 312 g of O₂ were obtained from the reaction.

Answer 2

Answer:

3.1 × 10² g

Explanation:

Step 1: Write the balanced decomposition reaction

KClO₃ ⇒ KCl + 1.5 O₂

Step 2: Calculate the moles of O₂ formed from 6.5 moles of KClO₃

The molar ratio of KClO₃ to O₂ is 1:1.5.

6.5 mol KClO₃ × 1.5 mol O₂/1 mol KClO₃ = 9.8 mol O₂

Step 3: Calculate the mass corresponding to 9.8 moles of O₂

The molar mass of O₂ is 32.00 g/mol.

9.8 mol × 32.00 g/mol = 3.1 × 10² g