Answer:
37.1 kilograms of ammonia gas will be produced in this process
Explanation:
The percentage yield of the reaction is given by:
According to question
The percentage yield of the given industrial process = 74.0%
The given theoretical yield of ammonia gas = 50.1 kg
The experimental yield of ammonia gas = x
The percentage yield of the reaction is calculated a:
Solving for x, we get:
37.1 kilograms of ammonia gas will be produced in this process
Parrallax is used to calculate the distance.
Answer:
0,72 moles of SO₂ remain
Explanation:
The reaction is:
2SO₂ + O₂ → 2SO₃
Where molecular mass of SO₂ is 64,066g/mol and of SO₃ is 80,066g/mol.
86,0g of SO₂ are:
86,0g × (1mol / 64,066g) = <em>1,34 moles of SO₂</em>.
50,0g of SO₃ are:
50,0g × (1mol / 80,066g) = <em>0,62 moles of SO₃</em>.
Now, as 2 moles of SO₂ produce 2 moles of SO₃, the moles of SO₂ that remain after the reaction are the initial moles of SO₂ - moles of SO₃:
1,34 moles - 0,62 moles =
<em>0,72 moles of SO₂ remain</em>
I hope it helps!
Answer:
Kp is 0.228/atm
Explanation:
This is the reaction:
CO + Cl<u>₂</u> → CCl₂O
1 mol of carbon monoxide and 1 mol of chlorine produce 1 mol of phosgene.
Formula for Kp which derivates from Kc is:
Kp = Kc (R.T)ⁿᵇ ⁻ ⁿᵃ
Δп = nb (moles in the products) - nₐ(moles in the reactants)
Δп = 1 - 2 = -1
T is T° in K → T°C + 273 = 611°C +273 = 884K
R → Universal constant gas → 0.082 L.atm/mol.K
We replace the data: Kp = 16.5 L/mol (0.082 . 884K)⁻¹ → 0.228/atm
