Moles of ammonia
- Given mass/Molar mass
- 230/17
- 13.5mol
1mol requires 5.66KJ energy
13.5mol requires
Answer:
n = 2.1 mol
Explanation:
Given data:
Number of moles of gas = ?
Volume of gas = 56.3 L
Pressure of gas = 0.899 atm
Temperature of gas = 20°C (20+273 = 293 k)
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
0.899 atm × 56.3 L = n × 0.0821 atm.L/ mol.K × 293 k
50.614 atm.L = n × 24.055 atm.L/ mol
n = 50.614 atm.L / 24.055 atm.L/ mol
n = 2.1 mol
Answer:
Mass = 24.4 g
Explanation:
Given data:
Mass of lithium nitride needed = ?
Volume of ammonia = 15.0 L
Solution:
Chemical equation:
Li₃N + 3H₂O → NH₃ + 3LiOH
One mole = 22.414 L
We will compare the volume of lithium nitride and ammonia.
NH₃ : Li₃N
22.414 : 22.414
15.0 : 15.0
Number of moles of lithium nitride:
PV = nRT
n = 1 atm ×15 L / 0.0821 atm.L / mol.K × 273 K
n = 15 /22.41 /mol
n =0.7
Mass of Lithium nitride:
Mass = number of moles × molar mass
Mass = 0.7 mol × 34.83 g/mol
Mass = 24.4 g