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

false

because it does not have the potential to be entirely used up

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )