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Evgesh-ka [11]
3 years ago
15

In an experiment, 200.00 grams of Al2O3 was decomposed producing 77.18 grams of Al. Calculate the percent yield for this experim

ent?
Chemistry
2 answers:
lakkis [162]3 years ago
6 0

Answer:

approximately 106 grams

Explanation:

The molecular mass of Aluminum Oxide is approximately 102 grams per mole.

2×Al=

2×27=54

3×O=3×16=48

54+48=102

200

102

=1.96 moles

There are two atoms of Aluminum in one mole and each atom has a mass of approximately 27 grams so

Mass of Al = 1.96×2×27=106 grams

Explanation:

zhuklara [117]3 years ago
4 0

Answer:

Explanation:

Molecular weight of Al: 27

Molecular weight of O: 16

Percentage composition of Al by weight in Al2O3

= (27*2) / (27*2 + 16*3)

= 0.5294

In 200 grams of Al2O3, there are

= 200*0.5294

= 105.88 grams of Al

77.18 grams of Al was produced in the experiment.

Percent yield for the experiment

= product mass / reactant mass * 100%

= 77.18 / 105.88 * 100%

= 72.89%

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How many moles are in 7.2x10^15 atoms of Pb?​
fiasKO [112]
<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

<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] 7.2 × 10¹⁵ atoms Pb

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb

<u>Step 4: Check</u>

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

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

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3 years ago
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