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MrRissso [65]
3 years ago
10

Gaseous ethane CH3CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. What is the

theoretical yield of carbon dioxide formed from the reaction of 18.0g of ethane and 57.8g of oxygen gas?
Chemistry
1 answer:
ololo11 [35]3 years ago
8 0

Answer:

46.97g

Explanation:

Hello,

To find the theoretical yield, we must first of all get the equation of reaction right in order to know how the compounds combine together.

Equation of reaction;

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

From the equation of reaction,

2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂.

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Mass of CO₂ = 4 × 44 = 176g

Mass of C₂H₆ = 2 × 30 = 60g

Mass of O₂ = 7 × 32 = 224g

Therefore, 224g + 60g (O₂ + C₂H₆) = 176g of CO₂.

284g of reactants = 176g of product(CO₂)

(18 + 57.8)g of reactants = x g of products(CO₂)

X = (75.8 × 176) /284

X = 46.97g of CO₂.

Theoretical yield of CO₂ is 46.97g

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General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

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<u>Stoichiometry</u>

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Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

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

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

24.9235 L Ar ≈ 24.9 L Ar

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