Answer:
2 mol H₂O
Explanation:
With the reaction,
1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:
Answer:
pH of a 0,245 M ammonia solution is 11,3 and percent ionization is 0,86%
Explanation:
For the equilibrium buffer of NH₃:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻; kb = 1,8x10⁻⁵
kb = [NH₄⁺] [OH⁻] / [NH₃] <em>(1)</em>
When 0,245 M of NH₃ is added, the equilibrium concentrations are:
[NH₃] = 0,245 - x
[NH₄⁺] = x
[OH⁻] = x.
Replacing this values in (1)
x² + 1,8x10⁻⁵x - 4,41x10⁻⁶ = 0
Solving for x:
x = -0,00211 No physical sense. There are not negative concentrations.
x = 0,00211 Real answer
Thus [OH⁻] in equilibrium is 0,00211 M.
As pOH = -log [OH⁻] and 14 = pH + pOH
pH of 0,00211 M is <em>11,3</em>
It is possible to calculate the percent ionization thus:
Percent ionization = [OH−] equilibrium / [B] initial×100%
Replacing:
0,00211 / 0,245 × 100 = <em>0,86%</em>
I hope it helps!