How many mL of O2 at STP are produced from 1.5 g KClO3

64.2 L

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Explanation:

In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.

Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.

STP conditions are defined as a pressure of

100 kPa

and a temperature of 0∘C

. Under these conditions for pressure and temperature, one mole of any ideal gas occupies

22.7 L

- this is known as the molar volume of a gas at STP.

So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.

The balanced chemical equation for this decomposition reaction looks like this

2KClO3(s]heat×−−−→2KCl(s]+3O2(g]↑⏐

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.