Question

Nitrogen gas reacts with hydrogen gas to produce ammonia. N2(g) + 3H2(g) — 2NH3(g) a. If you have 3.64 g of H, how many grams of NH3 can be produced? b. How many grams of H, are needed to react with 2.80 g of N? c. How many grams of NH; can be produced from 12.0 g of H? DO

9 please.

Answer 1

Dear student,

Moles of the produce = 2 x 1.82 mol Elo 21 mol Now, Mass of Imole of NH3 is 17g. Mass g 1.21 mole of NH3 is 1.21x 17 g = 20.579 oo Mass of NH3 produced - 201599

16 Mass q N2 = 289 Molar mass of Ng = 289/onal Moles of N2 = Mass of N2 Holay mass of Na = 2184 = 0.1 moli 2881mol from the reaction we conclude that for I mole of Ne, 3 mole og Hy is needed for ol mole of N2, 3X01 mole of H₂ is needed Holes of H₂ = 3x0,1 mol = 0.3 mol. Now, Mass & Imole q Ma is 29 Mass of 0.3 mole of H₂ is 0.3x2 g = 0.69 . Mass of the needed to react with 2.89 g N2 is 0.6g. (c) Mass q H₂ = 12g. Holes of H₂ = mass of H2 Molar Mass of H₂ = 12g = 6 mol. 2g/mol

from the reaction we conclude that for 3 mole of H₂ , 2 mole of NH3 produced for 6 mole of M2 , 2 x 6 mole of NH3 prodenced Moles of NH₃ produced = 2x6 mol = 4 not Now, mass of Imole of NH3 is 179 Mass of yonole of NH3 is 4x 17 g = 689 8 68 g q NH₂ will produced from 129 q H2.

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Answer 2

(A)1.21333333 grams

(B) 0.6 grams

(C)68 grams

Answer 3

(A)1.21333333 grams

(B) 0.6 grams

(C)68 grams

Answer 4

(A)1.21333333 grams

(B) 0.6 grams

(C)68 grams