Grade 12Physical Chemistry

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g) 2NH3 (g) (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

sudhanshu

12 Years agoGrade 12

5 Answers

Gaurav

11 Years ago

Hello Student

(i) Balancing the given chemical equation,

From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of
dihydrogen to give 2 mole (34 g) of ammonia.
⇒ 2.00 × 103g of dinitrogen will react with dihydrogen i.e.,2.00 × 103g of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 10³g
Hence, N2 is the limiting reagent.
28 g of N2 produces 34 g of NH3.
Hence, mass of ammonia produced by 2000 g of N2= 2428.57 g

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 willremain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 10³g – 428.6 g= 571.4 g

Amrutha amruzz

8 Years ago

in the reaction A+B gives AB2

identify the limiting reagent in the following

1.300atom of A+200molecules of B

2.2moles of A+3moles of B

3.100atom of A+100molecules of B

4. 5mol pf A+2.5mol of B

5.2.5mol of A+5mol of B

Jeel

7 Years ago

(i) According to the reaction, 1 atom of A reacts with 1 molecule of B.

∴200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B.

∴ 2 mol of A will react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting reagent.

(iii) 1 atom of A combines with 1 molecule of B.

∴ All 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric and ther is no limiting reagent.

(iv) 1 mol of atom A combines with 1 mol of molecule B.

∴ 2.5 mol of B will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence, B is the limiting reagent.

(v) 1 mol of atom A combines with 1 mol of molecule B.

∴ 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left. Hence, A is the limiting reagent.

Ajeet Tiwari

6 Years ago

hello students

(i) Balancing the given chemical equation, From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia. ⇒ 2.00 × 103g of dinitrogen will react with dihydrogen i.e.,2.00 × 103g of dinitrogen will react with 428.6 g of dihydrogen. Given, Amount of dihydrogen = 1.00 × 103g Hence, N2 is the limiting reagent. 28 g of N2 produces 34 g of NH3. Hence, mass of ammonia produced by 2000 g of N2= 2428.57 g

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 willremain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 103g – 428.6 g= 571.4 g

Hope it helps
Thankyou

Yash Chourasiya

5 Years ago

Dear Student

(i) Balancing the given chemical equation,
From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.
⇒ 2.00 × 103g of dinitrogen will react with dihydrogen i.e.,2.00 × 103g of dinitrogen will react with 428.6 g of dihydrogen.
Given, Amount of dihydrogen = 1.00 × 103g
Hence, N₂ is the limiting reagent. 28 g of N₂ produces 34 g of NH₃.
Hence, mass of ammonia produced by 2000 g of N₂ = 2428.57 g

(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ willremain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 103g – 428.6 g= 571.4

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya