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Advocard [28]
3 years ago
8

At 506 K and 2.05 atm, if 6.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3? (R = 0.0820

6 L atm/mol K)
Chemistry
1 answer:
Nady [450]3 years ago
8 0

Answer:

  • 90.4 liter

Explanation:

You must convert the 6.00 grams of H₂ into number of moles, and then use the stoichiometry of the reaction to find the number of moles of NH₃, which can be converted into volume using the ideal gas equation.

<u></u>

<u>1. Number of moles of H</u><u>₂</u>

  • number of moles = mass in grams / molar mass
  • molar mass of H₂ = 2.016g/mol
  • number of moles = 6.00grams / 2.016g/mol = 2.97619mol

<u></u>

<u>2. Number of moles of NH</u><u>₃</u>

<u></u>

i) Chemical equation:

  • 3H₂(g) + N₂(g) →  2NH₃(g)

ii) Mole ratio:

  • 3 mol H₂ : 2 mol NH₃

iii) Proportion:

  • x mol  NH₃ / 2.97619mol H₂ = 3 mol NH₃ / 2 mol H₂

  • x = 4.4642857mol NH₃

<u>3. Volume of NH₃</u>

  • pV = nRT
  • V = nRT/p
  • V = 4.4642857mol × 0.08206 atm·liter/(K·mol) × 506K / 2.05atm
  • V = 90.4 liter
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