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Elan Coil [88]
3 years ago
13

The theoretical yield of ammonia in an industrial synthesis was 550 kg, but only 480 kg was obtained. What was the percentage yi

eld of the reaction?
Chemistry
1 answer:
bezimeni [28]3 years ago
8 0

Theoretical yield 550 kg - 100%

Actual yield 480 kg - x%


x= 480*100/550 ≈ 87.3 %


Percentage yield ≈ 87.3 %.



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<h3>Answer:</h3>

28 mol CaF

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                          \displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 28.2298 \ moles \ CaF

<u>Step 4: Check</u>

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

28.2298 mol CaF ≈ 28 mol CaF

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