Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
Umm...Well...
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The volume of O₂ produced: 84.6 L
<h3>Further explanation</h3>
Given
7.93 mol of dinitrogen pentoxide
T = 48 + 273 = 321 K
P = 125 kPa = 1,23365 atm
Required
Volume of O₂
Solution
Decomposition reaction of dinitrogen pentoxide
2N₂O₅(g)→4NO₂(g)+O₂ (g)
From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :
= 0.5 x mol N₂O₅
= 0.5 x 7.93
= 3.965 moles
The volume of O₂ :
Is this a multi choice question?????????????????????
