Answer:
6.142 moles of NaCl
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2AlCl3 + 3Na2S —> Al2S3 + 6NaCl
Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:
Mass mass of Na2S = 78.048g/mol
Mass of Na2S = 239.7g
Number of mole Na2S =..?
Mole = Mass /Molar Mass
Number of mole Na2S = 239.7/78.048 = 3.071 moles
Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:
From the balanced equation above,
3 moles of Na2S reacted to produce 6 moles of NaCl.
Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl
Flammability
Hope it helps!
Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
