Magnesium oxide : MgO
Calcium bicarbonate: Ca(HCO3)2
aluminum carbonate: Al2(CO3)3 or C3Al2O9
Answer:
43.05 moles of Al needed to react with 28.7 moles of FeO.
Explanation:
Given data:
Moles of FeO = 28.7 mol
Moles of Al needed to react with FeO = ?
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Now we will compare the moles of Al with FeO.
FeO : Al
2 : 3
28.7 : 3/2×28.7 = 43.05 mol
Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.
<h3>
Answer:</h3>
25.4 g CH₄
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.58 mol CH₄
<u>Step 2: Identify Conversions</u>
[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: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
25.359 g CH₄ ≈ 25.4 g CH₄
Answer:
option B
cementation and compaction