Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
The reaction corresponds to the combustion of propane (C3H8). The balanced reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
The reaction enthalpy is given as:
ΔHrxn = ∑nΔH°f(products) - ∑ nΔH°f(reactants)
= [3ΔH°f(CO2(g)) + 4ΔH°f(H2O(g)] - [1ΔH°f(C3H8(g)) + 5ΔH°f(O2(g)]
= [3(-393.5) + 4(-241.8)] - [-103.9 + 5(0)] = -2043.8 kJ
The enthalpy for the combustion of propane is -2044 kJ
<h3>
Answer:</h3>
2265 g Fe₃O₄
<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>
[RxN - Balanced] Fe₃O₄ + 4H₂ → 3Fe + 4H₂O
[Given] 705.0 g H₂O
<u>Step 2: Identify Conversions</u>
[RxN] 4 mol H₂O → 1 mol Fe₃O₄
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Fe - 55.85 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
Molar Mass of Fe₃O₄ - 3(55.85) + 4(16.00) = 231.55 g/mol
<u>Step 3: Convert</u>
- Set up stoich:

- Multiply/Divide/Cancel units:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
2264.74 g Fe₃O₄ ≈ 2265 g Fe₃O₄