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
The answer is statement #3.
Answer:
4 × 10 g
Explanation:
Step 1: Write the balanced equation
2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)
Step 2: Calculate the moles corresponding to 4 g of H₂
The molar mass of H₂ is 2.02 g/mol.
4 g × 1 mol/2.02 g = 2 mol
Step 3: Calculate the moles of H₂O produced from 2 moles of H₂
The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.
Step 4: Calculate the mass corresponding to 2 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
2 mol × 18.02 g/mol = 4 × 10 g
If the temperature of a liquid-vapor system at equilibrium increases, it will shift towards the vapor phase, assuming that the pressure remains equal. The concentration of vapor will also increase relative to the concentration of liquid in the system. Thus, the new equilibrium condition will have more vapor than liquid.
The answer is 6.022• 10^23 atoms