<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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>
- Reading a Periodic Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [DA} Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40
Answer:
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