2KCIO3 > 2KCI + 302
How many grams of O2 will be formed from 6.7 grams of KCIO3?

2.6253 g O2 formed  

Explanation:

Given 6.7 moles of KClO3 as the target amount in the reactant, used the coefficient of the balanced chemical reaction involved to determine the number of grams of O2 molecules formed.  

x g O2 = 6.7 g KClO3 x [(1 mol KClO3)/ (122.5 g KClO3)] x [(3 mol O2)/ (2 mol KClO3)] x [(32 g O2)/ (1 mol O2)]  

x g O2  = 2.6253 g O2 formed  

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.