Answer:The product and reactants reach a final, unchanging level.
If reactants eventually collide,
there is an occurrence of reaction.
<span>
Therefore, when there is an increase concentration of
reactant, meaning to say that there are several moles of it every unit volume. An
example of this is a room having hundred of people will absolutely get higher
concentration compared to a room with one individual only.
Pertaining to effective collisions, if ever there is an
increase of concentration, the frequency and rate of effective collisions among
reactants surges in such a way that the rate of reaction also surges. Same with
passing into a room with only 1 individual compared to hundred people blind
persons, you probably want to proceed to the room with several people.</span>
<span>This is the simple logic
behind that scientific existence.</span>
Molar mass:
H₂O = 18.0 g/mol
O₂ = 32.0 g/mol
C₅H₁₂ + 8 O₂ -> 5 CO₂ + 6 H₂<span>O
</span>
8 x (32 g )<span> ------------ 6 x (18 g )</span>
mass O₂ ------------ 108 g H₂O
mass O₂ = 108 x 8 x 32 / 6 x 18
mass O₂ = 27648 / 108
mass O₂ =<span> 256 g</span>
<span>hope this helps!</span>
Answer:
24x10³
Explanation:
2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)
The equilibrium constant for this reaction is:
Kc = ![\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BO_2%5D%5E3%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E4%7D)
The expression of [CH₃OH] is left out as it is a pure liquid.
Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:
- CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
- H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
- O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂
Then we calculate the concentrations:
- [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
- [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
- [O₂] = 0.0875 mol / 7.5 L = 0.0117 M
Finally we <u>calculate Kc</u>:
- Kc =
= 24x10³
