Answer:
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
Answer:
Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm
3
of oxygen at STP
