Answer:
1. 48 mols
2. 0.2 M
5. 1.25 L
Explanation:
Molarity= mols divided by liters
Hope this helps not sure about 3 and 4
Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Answer:
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Explanation:
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
No, they do not. Hope I helped! :)
