CHEMISTRY | QN 02 | MOLE CONCEPT
Hydrogen gas reached with nitrogen gas to form ammonia by Haber process. Calculate the volume of ammonia produced of 2dm3 of nitrogen reacted with 4dm3 of hydrogen gas at S.T.P.
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Data
Volume of N2 = 2dm3
Volume of hydrogen (H2) = 4dm3
Molar volume at S.T.P = 22.4dm3
Asked: to calculate the volume of Ammonia gas produced -
CHEMICAL EQUATION
N2(g) + 3H2(g) ⇋ 2NH2
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To convert volume of N2 and H2 into moles.
Mole = volume / Molar volume
For N2 = 2dm3 / 22.4 dm3/mole
= 0.089 moles
Mole ratio for N2 = mole calculated / Mole from equation
= 0.089 moles / 1mole
= 0.089
Mole ratio for H2 = 4dm3 / 22.4 dm3/mole
= 0.17857 mole
Mole ratio for H2 = 0.17857 mole / 3
= 0.0595 -
Mole equivalent between hydrogen gas and ammonia gas
3moles of H2 Ξ 2moles of NH3
0.17857 mole Ξ??
> 3 moles × x = 0.17857 moles × 2 moles
= 0.17857 × 2 3 = 0.11904 mole of NH3
Volume = mole × molar volume
= 0.11904 mole × 22.4dm3
= 2.67dm3
2.67dm3 of ammonia gas was produced.
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Prepared By:
Sir Mahame
Biology & Chemistry teacher,
Mapambano Education Center | Mwenge | DSM.
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