Answer: Option (4) is the correct answer.
Explanation:
It is known that equilibrium constant is represented as follows for any general reaction.

K = ![\frac{[C][D]}{[A][B]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BC%5D%5BD%5D%7D%7B%5BA%5D%5BB%5D%7D)
As equilibrium constant is directly proportional to the concentration of products so more is the value of equilibrium constant more will be the number of products formed.
As a result, more is the time taken by the reaction to reach towards equilibrium. Whereas smaller is the value of equilibrium constant more rapidly it will reach towards the equilibrium.
Thus, we can conclude that cases where K is a very small number will require the LEAST time to arrive at equilibrium.
Answer:
1.15 atm
Explanation:
According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.
Therefore we have:
Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide
We were given the following:
Total pressure = 2.45 atm
Pressure of oxygen = 0.65 atm
Pressure of carbon monoxide = x
Pressure of carbon dioxide = 0.65 atm
Therefore:
2.45 = x + 0.65 + 0.65
2.45 = x + 1.3
x = 2.45 - 1.3
x = 1.15 atm
2Al + 3Cl₂ → 2AlCl₃
mol Al = 2/3 x 1.25 = 0.83
mass Al = 0.83 x 27 g/mol = 22.41 g
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂