Question

Find the volume of oxygen produced by 1mole of potassium trioxochlorate(v) at stp in the following reaction
2KCLO3➡️2KCL +3O2

Answer 1

According to the balanced chemical equation, 2 moles of potassium chlorate (KCLO3) react to produce 3 moles of oxygen gas (O2).

Therefore, for every 2 moles of potassium chlorate, 3 moles of oxygen gas are produced.

Since we want to find the volume of oxygen gas produced by 1 mole of potassium chlorate, we need to calculate the molar volume of oxygen gas at STP (Standard Temperature and Pressure).

The molar volume of a gas at STP is approximately 22.4 liters/mol.

Therefore, if 2 moles of potassium chlorate produce 3 moles of oxygen gas, then 1 mole of potassium chlorate will produce (3/2) moles of oxygen gas.

To find the volume of oxygen gas produced by 1 mole of potassium chlorate, we multiply the number of moles by the molar volume:

Volume = (3/2) moles * 22.4 liters/mol = 33.6 liters

Thus, the volume of oxygen gas produced by 1 mole of potassium trioxochlorate(V) at STP is 33.6 liters.