Answer :
(a) The mass of
produced is, 15.2 grams.
(b) The percent yield of the reaction is, 72.5 %
Explanation :
Part (a) :
Given,
Mass of
= 85.1 g
Molar mass of
= 27 g/mol
First we have to calculate the moles of 

Now we have to calculate the moles of 
The balanced chemical equation is:

From the reaction, we conclude that
As, 4 moles of
react to give 2 moles of 
So, 3.15 moles of
react to give
mole of 
Now we have to calculate the mass of 

Molar mass of
= 102 g/mole

Therefore, the mass of
produced is, 161.2 grams.
Part (b) :
Now we have to calculate the percent yield of the reaction.

Experimental yield = 116.9 g
Theoretical yield = 161.2 g
Now put all the given values in this formula, we get:

Therefore, the percent yield of the reaction is, 72.5 %