If 15.7 grams of A1203 are produced, how many grams of iron (Fe) were

1 answer:

4 0

Answer:

77.4 g Fe

Explanation:

g Al2O3 —— mol Al2O3 —— mol Fe —- g Fe

15.7 g Al2O3 x (1 mol Al2O3/101.959 g Al2O3) x (3 mol Fe/1 mol Al2O3) x (167.535 g Fe/1 mol Fe) = 77.4 g Fe

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Determine the number of grams of carbon 3.14x 1023 atoms of carbon

tiny-mole [99]

<h3>Answer:</h3>

6.26 g C

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
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] 3.14 × 10²³ atoms C

[Solve] grams C

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

Avogadro's Number

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

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 3.14 \cdot 10^{23} \ atoms \ C(\frac{1 \ mol \ C}{6.022 \cdot 10^{23} \ atoms \ C})(\frac{12.01 \ g \ C}{1 \ mol \ C})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 6.26227 \ g \ C$