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Directions: Solve the following problems. Write your complete solution and answers on a separate sheet of paper. 1. The fertilizer ammonium sulfate is prepared by the reaction between ammonia and sulfuric acid. (Molar Mass: N=14, S=32, H=1, O=16) 2NH3(g) + H2S04(aq) → (NH4)2S04 a. What is the molar mass of NH3? b. What is the molar mass of (NH4)2SO4? c. What is the mole-mole ratio of NH3 to (NH4)2SO4? d. How many moles of ammonium sulfate are produced when 638.4 g of NH reacts completely with H2SO4? e. How many grams of NH3 are needed to produce 50.0 g of (NH4)2SO4? 2. In the fermentation process, ethanol is produced from the decomposition glucose. (Molar mass: C=12, H=1, O=16) CAH,0a C,H,OH + CO2 206​

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CBSE

Chemistry

Grade 11

Chemical Reactions

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Ammonia reacts with sulfuric acid to give the fertilizer ammonium sulphate. Calculate the volume of ammonia atSTPatSTP used to form 59 g of ammonium sulphate.

N=14,H=1,S=32,O=16N=14,H=1,S=32,O=16

A- 20.02 L

B-24.33 L

C-66.20 L

D-34.12 L

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Hint: First of all, write the balanced chemical equation for the reaction of ammonia and sulfuric acid then apply volume of compound as per mole concept (i.e. 22.4 L volume for one mole of compound).

Complete Solution :

Firstly, write the balanced chemical equation for the given reaction.

Ammonia + Sulphuric acid → Ammonium sulphate

Ammonia + Sulphuric acid → Ammonium sulphate

i.e. 2NH3+H2SO4→(NH4)2SO4

2NH3+H2SO4→(NH4)2SO4

At S.T.P. i.e. Standard temperature and pressure condition volume of one mole of gas is 22.4 L.

-Volume of one mole of ammonia = 22.4 L

-Volume of two moles of ammonia is calculated as,

= 22.4 x 222.4 x 2

= 44.8L44.8L

(As 2 moles of ammonia are required for formation of one mole ammonium sulphate.)

- Molecular weight of ammonium sulphate is calculated as,

= (14 x 2) + (1 x 8) + (32 x 1) + (16 x 4)

(14 x 2) + (1 x 8) + (32 x 1) + (16 x 4)

= 132 g

Volume of ammonia required to form 132g ammonium sulphate = 44.8 L

Volume of Ammonia required to form 59 g ammonium sulphate is calculated as,

= 44.8132×5944.8132×59

= 20.02 L

From the above statement we can conclude that the volume of ammonia required used to form 59g of ammonium sulphate is 20.02 L.