Di nitrogen and di hydrogen reacts with each other to prduce ammonia a

1.	Di nitrogen and di hydrogen reacts with each other to prduce ammonia according to reactionN2(g) +3H2(g) ----- 2NH3(g)1)calculate the mass of ammonia if 2x10^3g of di nitrogen reacts with 1x10^3g of di hydrogen?2)will any of the 2 reaction remains unreacted ?3)if yes,which one and wat would b its mass
Answer» (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

Di nitrogen and di hydrogen reacts with each other to prduce ammonia according to reactionN2(g) +3H2(g) ----- 2NH3(g)1)calculate the mass of ammonia if 2x10^3g of di nitrogen reacts with 1x10^3g of di hydrogen?2)will any of the 2 reaction remains unreacted ?3)if yes,which one and wat would b its mass

Answer»

(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