Determine the formula mass of KClO3

Decomposition of potassium chlorate yields potassium chloride and oxygen as:

2KClO

3

→2KCl+3O

2

Thus 2 moles of potassium chlorate yields 3 moles of oxygen gas.

2 moles of potassium chlorate =2×122.5=245g of potassium chlorate

At STP, the volume occupied by 1 mol of gas =22.4 dm

3

the volume occupied by three moles of a gas =3×22.4=67.2dm

3

Therefore, 245g of potassium chlorate yields 67.2dm

3

of oxygen gas

To liberate 6.72 dm

3

oxygen amount of potassium chlorate required is

=

67.2

245

×6.72=24.5g

Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm

3

of oxygen at STP

Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm

3

of oxygen at STP

the answer is 303.15 or 303

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be