What is the limiting reactant if 70.0 g of c7h6o3 and 80.0 g of c4h6o3 react?

Given:

70g C7H6O3

80g C4H6O3

Required:

Limiting reactant

Solution:

The balanced chemical reaction is C7H6O3 + C4H6O3 + 23/4O2 -> 11CO2 + 6H2O

Molar mass of C7H6O3 = 138 g/mol

Molar mass of C4H6O3 = 102 g/mol

70g C7H6O3 (1 mol C7H6O3/138g C7H6O3)(1 mol C4H6O3/ 1mol C7H6O3)(102g C4H6O3/1mol C4H6O3) = 51.74g C4H6O3

80g C4H6O3 - 51.74g C4H6O3 = 28.26g C4H6O3

There are 28.26g C4H6O3 left after reacting it with C7H6O3, therefore the limiting reactant is C7H6O3.

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.