Question

How many nuclear protons are in a molar quantity of `0.667*mol` with respect to `KClO_3`?

Answer 1

In `"potassium chlorate"`, I get `0.667*mol....` with respect to `KClO_3`...

Explanation:

To find a molar quantity of stuff we take the quotient....

`"mass of stuff"/"molar mass of stuff"=(81.7*g)/(122.55*g*mol^-1)=0.667*mol`...

Now each mole of `KClO_3` contains `19_K+17_"Cl"+3xx8_"O"=60*"nucular protons"xxN_A`, where `N_A-="Avogadro's Number"`....and here note that we speak of nuclear protons, massive, positively charged, nucular particles....

And so in the given quantity we take the product....

`60*"protons"xxN_Axx0.667*mol^-1=40xxN_A`....or `40*mol` of protons as required.