4 NH3 +5 02 +4NO + 6H2O When 70.25 grams of NH3 react, how many moles of H20 are produced?
1 answer:
Answer:
6.19 mol H₂O
Explanation:
- 4 NH₃ + 5O₂ → 4NO + 6H₂O
First, we<u> convert 70.25 grams of NH₃ into moles</u>, using its<em> molar mass</em>:
- 70.25 g ÷ 17 g/mol = 4.13 mol NH₃
Then we <u>convert 4.13 moles of NH₃ into moles of H₂O</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4.13 mol NH₃ *
= 6.19 mol H₂O
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Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0