1. The total moles of the solution is 0.3079193 mol.
2. The mole fraction for gold is 0.2473212, and for silver, it is 0.7526787.
3. The molar entropy of mixing for gold is 2.87285 j/K, while for silver, it is 1.77804 j/K.
4. The total entropy of mixing sums to 4.65089 j/K.
5. Molar free energy amounts to -2325.445 kJ.
6. Chemical potential for silver is -1750.31129 j/mol and for gold, it is -575.13185 j/mol. To elaborate:
(1) The molar mass of silver stands at 107.8682 g/mol, and gold's at 196.96657 g/mol. Hence, calculating moles leads to mass/molar mass for silver: 25 g/107.8682 g/mol = 0.2317643 mol and for gold: 15 g/196.96657 g/mol = 0.076155 mol, resulting in a total of 0.30791193 mol.
(2) For the mole fractions, silver's fraction is 0.2317643/0.3079193 = 0.7526787, and for gold, it's 0.076155/0.3079193 = 0.2473212.
(3) To find molar entropy mixing ∆Sm, we use the formula ∆Sm = -R * Xi * ln(Xi) where R = 8.3144598. For silver, substituting gives us 1.77804 j/K, while for gold, we get 2.87285 j/K.
(4) Overall entropy of mixing totals 4.65089 j/K thus calculated.
(5) The Gibbs free energy at 500 °C can be derived through G = H - TS, accounting to H = 0 (as T is 500 + 273 = 773 K and S is 4.65089), resulting in G equating to -3595.138 kJ.
(6) The chemical potentials calculated derive from multiplying the Gibbs free energy by their mole fractions.
The concentration of the HCl solution can be determined as follows:
The reaction equation is written as
NaOH + HCl = NaCl + H2O
Next, the moles of NaOH are calculated: moles = molarity x volume /1000
= 5 x 2/1000 = 0.01 moles
Using the mole ratio of NaOH to HCl, which is 1:1, the moles of HCl is also equal to 0.01 moles
The concentration is given by: concentration = moles/volume x 1000
= 0.01/10 x 1000 = 1M
Result:
94.7 %
Explanation:
The balanced reaction is:
2 S + 3 O₂ → 2 SO₃
The stoichiometric mole ratio is:
S: 2 moles
O₂: 3 moles
Moles are calculated as mass divided by molar mass:
n = w / m
where n = moles, w = mass, m = molar mass.
Given:
For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol
For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol
Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.
Using proportions:
3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.
Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.
Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g
The percentage yield is actual yield divided by theoretical yield times 100:
Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %
