Question

how many moles potassium chlorate is required to produce 4.8 gm of oxygen​

Answer 1

To Find -

• How many moles of KClO3 is required to produce 4.8 g of oxygen

Solution :-

• 2KCLO3 → 2KCl + 3O2

This equation shows that If we heat 2 moles of KClO3 then it gives 2 moles of KCl and 3 moles of O2.

Then,

Let's calculate the moles of 4.8 g of oxygen.

• mass = n × GAM

→ 4.8 = n × 32

→ n = 4.8/32

→ n = 3/20

→ n = 0.15 mol

Then,

2 moles of KClO3 produces 3 moles of O2.

Thus,

→ 2/3 × 0.15

→ 0.3/3

→ 0.1 mol

Hence,

0.1 mol of KClO3 is required to produce 4.8 g of O2.