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?
(i) Balancing the given chemical equation,
Total mass of Ammonia = 2((14) +3(1))= 34 g
From the chemical equation,we can write
28gm of N2 reacts with 6gm of H2 to produce ammonia= 34g
Or 1 gm of N2 reacts with 1gm of H2 to produce ammonia= 34/28*1
Or when 2.00x103 g of N2 reacts with 1.00x103 gm of H2 to produce ammonia
=34/28 *2.00x103=2428.57g
hence 2.00 × 103 g of dinitrogen will react with 1.00x103 g of dihydrogen to give 2428.57 g of ammonia
Given, Amount of dihydrogen = 1.00 × 103 g
Hence, N2 is the limiting reagent.
(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1.00 × 103 g – 428.6 g
= 571.4 g
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Wow! Gr8
Good explanation , easy to understand
Any easy method
Nice
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Bad explanation... It\'s shortcut
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