Question
Question: The percentage of N in 66 g pure \[{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\...
The percentage of N in 66 g pure (NH4)2SO4sample is :
A . 32
B . 28
C . 21
D . none of the above
Solution
(NH4)2SO4, Ammonium sulfate is an inorganic compound. To calculate the percentage at first you have to calculate the molecular weight of the molecule. Ammonium sulfate salt can be prepared from ammonia and sulfuric acid.
Complete step by step answer:
Ammonium sulfate can be prepared from the reaction between NH3 and H2SO4. In this reaction one mole of H2SO4 reacts with two moles of NH3. As ammonia is a base it absorbs hydrogen ion and forms ammonium ion. Then it combines with the sulfate anion to form the ionic inorganic salt ammonium sulfate. The reaction is shown below.
H2SO4 + 2NH3→2NH4 + + SO4 - 2→(NH4)2SO4
Now the molecular weight of ammonium sulfate,(NH4)2SO4 is.
When this salt is 100%pure, the percentage of nitrogen present is
,
Now, for 66 gram pure ammonium sulfate the percentage of nitrogen is
=1322×14×66 =13228×66 =14%This answer is not matching with any of those given answers in the question. So, none of those is the correct answer.
Therefore, the correct option is D.
Note:
Ammonium sulfate is hygroscopic in nature. For alkaline soils ammonium sulfate is used as a fertilizer, it releases ammonium ion into the soil and produces a small amount of acids to maintain the pH balance of the soil. Ammonium sulfate also contributes an essential nitrogen element into the soil for the growth of the crops.