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Question
a) Calculate the mass of ammonia produced if 2.00 ×103 g of nitrogen reacts with 1.00 ×103 g.
b) Will any of the two reagents will remain unreacted. If yes, which one and what would be its mass.
a) Calculate the mass of ammonia produced if 2.00 ×103 g of nitrogen reacts with 1.00 ×103 g.
b) Will any of the two reagents will remain unreacted. If yes, which one and what would be its mass.
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Solution
N2(g) + H2(g) → 2NH3(g)
Balancing the chemical equation, From the equation, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g) of ammonia. ⇒ 2.00 × 10^3g of N2 will react with H2 i.e.,2.00 × 10^3g of N2 will react with 428.6 g of H2 Given, Amount of H2= 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 N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 willremain unreacted. Mass of H2 left unreacted = 1.00 × 103g – 428.6 g= 571.4 g
N2(g) + H2(g) → 2NH3(g)
Balancing the chemical equation, From the equation, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g) of ammonia. ⇒ 2.00 × 10^3g of N2 will react with H2 i.e.,2.00 × 10^3g of N2 will react with 428.6 g of H2 Given, Amount of H2= 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 N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 willremain unreacted. Mass of H2 left unreacted = 1.00 × 103g – 428.6 g= 571.4 gSuggest Corrections
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