Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer:
Mass = 2.89 g
Explanation:
Given data:
Mass of NH₄Cl = 8.939 g
Mass of Ca(OH)₂ = 7.48 g
Mass of ammonia produced = ?
Solution:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
Number of moles of NH₄Cl:
Number of moles = mass/molar mass
Number of moles = 8.939 g / 53.5 g/mol
Number of moles = 0.17 mol
Number of moles of Ca(OH)₂ :
Number of moles = mass/molar mass
Number of moles = 7.48 g / 74.1 g/mol
Number of moles = 0.10 mol
Now we will compare the moles of ammonia with both reactant.
NH₄Cl : NH₃
2 : 2
0.17 : 0.17
Ca(OH)₂ : NH₃
1 : 2
0.10 : 2/1×0.10 = 0.2 mol
Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.17 mol × 17 g/mol
Mass = 2.89 g
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
![Ksp=[Ag^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%2B%5D%5BCl%5E-%5D)
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
Answer:
6.67 moles
Explanation:
Given that:-
Moles of hydrogen gas produced = 10.0 moles
According the reaction shown below:-

3 moles of hydrogen gas are produced when 2 moles of aluminium undergoes reaction.
Also,
1 mole of hydrogen gas are produced when
moles of aluminium undergoes reaction.
So,
10.0 moles of hydrogen gas are produced when
moles of aluminium undergoes reaction.
<u>Moles of Al needed =
moles = 6.67 moles</u>
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