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Maurinko [17]
3 years ago
7

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o

f the work used to solve this problem.
2 K + F2 ---> 2 KF
Chemistry
1 answer:
Degger [83]3 years ago
8 0
1) Molar mass:

K = <span> 39.0983 g/mol
F</span>₂ = <span> 37.99 g/mol

moles ratio:

2 K + F</span>₂<span>   =   2 KF

2 x 39.0983 g K ----------------- 37.99 g F</span>₂
23.5 g K -------------------------- ( mass of F₂)
<span>
mass of F</span>₂ = 23.5 x 37.99 / 2 x 39.0983
<span>
mass of F</span>₂ = 892.765 / 78.1966
<span>
mass of F</span>₂ = 11.4169 g
<span>
Therefore:

</span>37.99 g F₂------------------ 22.4 L ( at STP)
11.4169 g F₂ --------------- ( volume F₂ )

Volume F₂ = 11.4169 x 22.4 / 37.99

Volume F₂ = 255.73856 / 37.99

Volume F₂ = 6.731 L
<span>


</span>
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CH4 + O2 → CO2 + H2O
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Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Organic</u>

  • Naming carbons

<u>Stoichiometry</u>

  • Analyzing reaction rxn
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 130 g CH₄

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

[PT] Molar Mass of H: 1.01 g/mol

Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                   \displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

8 0
3 years ago
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