Answer :
Part 1 : Balanced reaction, 
Part 2 : The theoretical yield of
gas = 440.96 g
Part 3 : The % yield of ammonia is 90.03 %
Solution : Given,
Mass of
= 475 g
Molar mass of
= 28 g/mole
Molar mass of
= 17 g/mole
Experimental yield of
= 397 g
<u>Answer for Part (1) :</u>
The balanced chemical reaction is,

<u>Answer for Part (2) :</u>
First we have to calculate the moles of
.

From the given reaction, we conclude that
1 moles of
gas react to give 2 moles of
gas
16.96 moles of
gas react to give
moles of
gas
Now we have to calculate the mass of
gas.


Therefore, the theoretical yield of
gas = 440.96 g
<u>Answer for Part (3) :</u>
Formula used for percent yield :


Therefore, the % yield of ammonia is 90.03 %