According to the equation above, how many moles of potassium chlorate, kclo3, must be decomposed to generate 1.0 l of o2 gas at standard temperature and pressure?

Moles of potassium chlorate = 0.02976 moles

Explanation:

At standard pressure and temperature,

22.4 L of a gas consists of 1 mole

Thus, given, volume of = 1.0 L

So,

1 L of a gas consists of mole

Moles of oxygen gas = 0.04464 moles

The reaction is shown below as:-

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

1 mole of oxygen gas are produced when moles of potassium chlorate undergoes reaction.

Thus,

0.04464 mole of oxygen gas are produced when moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate = 0.02976 moles

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

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