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katrin2010 [14]
2 years ago
6

Consider the chemical equation. 2NBr3 + 3NaOH Right arrow. N2 + 3NaBr + 3HOBr If there are 40 mol of NBr3 and 48 mol of NaOH, wh

at is the excess reactant?
Chemistry
1 answer:
Maslowich2 years ago
4 0

Answer:

NBr₃ is the excess reactant.

Explanation:

  • 2NBr₃ + 3NaOH → N₂ + 3NaBr + 3HOBr

We can solve this problem by <u>calculating how many moles of NaOH would react completely with 40 moles of NBr₃</u>, using the <em>stoichiometric coefficients </em>of the reaction:

  • 40 mol NBr₃ * \frac{3molNaOH}{2molNBr_3} =60 moles NaOH

As the required number of NaOH moles is higher than the available number (60 required vs 48 available), NaOH is the limiting reactant.

As such, the excess reactant is NBr₃.

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Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 41
RSB [31]

Answer:

pH → 7.47

Explanation:

Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.

Caffeine + H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

1 mol of caffeine in water can give hydroxides and protonated caffeine.

We convert the concentration from mg/L to M

415 mg = 0.415 g

0.415 g / 194.19 g/mol = 2.14×10⁻³ mol

[Caffeine] = 2.14×10⁻³  M

Let's calculate pH. As we don't have Kb, we can obtain it from pKb.

- log Kb = pKb → 10^-pKb = Kb

10⁻¹⁰'⁴ = 3.98×10⁻¹¹

We go to equilibrium:

Caffeine + H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)

X will be the amount of protonated caffeine and OH⁻

     Caffeine     +    H₂O  ⇄  Caffeine⁺  +  OH⁻      Kb

   (2.14×10⁻³ - x)                         x                x

We make the expression for Kb:

3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)

We can missed the -x in denominator, because Kb it's a very small value.

So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³

√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷

That's the [OH⁻].  - log [OH⁻] = pOH

- log 2.92×10⁻⁷ = 6.53 → pOH

14 - pOH = pH → 14 - 6.53 = 7.47

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