Question

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 = 0.08206 L atm/mol K)

Answer 1

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.

1. Number of moles of H₂

• 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

2. Number of moles of NH₃

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

3. Volume of NH₃

• pV = nRT

• V = nRT/p

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

• V = 90.4 liter